Class 11

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.4 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.4 Questions And Answers Maharashtra Board

Question 1.
There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and marble is picked from the chosen bag. What is the probability that the chosen marble is red?
Solution:
Let event R: Chosen marble is red.
Let event B_i: ith bag is chosen.
∴ P(B_i) = \(\frac{1}{3}\)
If Bag 1 is chosen, it has 75 red and 25 blue marbles.
∴ Probability that the chosen marble is red under the condition that it is from Bag 1 = P(R/B₁)
= \(\frac{{ }^{75} \mathrm{C}_{1}}{{ }^{100} \mathrm{C}_{1}}\)
= \(\frac{75}{100}\)
= 0.75
Similarly we get,
P(R/B₂) = \(\frac{60}{100}\) = 0.60
P(R/B₃) = \(\frac{45}{100}\) = 0.45
∴ Required probability
P(R) = P(B₁) P(R/B₁) + P(B₂) P(R/B₂) + P(B₃) P(R/B₃)
= \(\frac{1}{3}\)(0.75) + \(\frac{1}{3}\)(0.60) + \(\frac{1}{3}\)(0.45)
= \(\frac{1}{3}\)(1.8)
= 0.60

Question 2.
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from
(i) first box
(ii) second box
Solution:
Let event A₁: The ball is drawn from 1st box and
event A₂: The ball is drawn from the 2nd box.
∴ P(A₁) = \(\frac{1}{2}\), P(A₂) = \(\frac{1}{2}\)
Let event B: The ball drawn is pink.
There are 5 balls in the 1st box, of which 3 are pink.
∴ P(B/A₁) = \(\frac{3}{5}\)
There are 9 balls in the 2nd box, of which 5 are pink.
∴ P(B/A₂) = \(\frac{5}{9}\)
(i) By Bayes’ theorem,
the probability that a pink ball is drawn from the first box, is given by

(ii) By Bayes’ theorem,
the probability that a pink ball is drawn from the second box, is given by

Question 3.
There is a working women’s hostel in a town, where 75% are from neighbouring town. The rest all are from the same town. 48% of women who hail from the same town are graduates and 83% of the women who have come from the neighbouring town are also graduates. Find the probability that a woman selected at random is a graduate from the same town.
Solution:
Let the total number of women be 100.
∴ n(S) = 100
Let event N: Women are from neighbouring town,
event W: Women are from same town and
event G: Women are graduates.
Number of women from neighbouring town,
n(N) = 75
Number of women from same town,
n(W) = 25
∴ P(N) = \(\frac{n(N)}{n(S)}=\frac{75}{100}\) and
P(W) = \(\frac{n(W)}{n(S)}=\frac{25}{100}\)
P(G/N), P(G/W) represent probabilities that woman is graduate given that she is from neighbouring town or same town respectively.
∴ P(G/N) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{N})}{\mathrm{n}(\mathrm{S})}=\frac{83}{100}\) and
P(G/W) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{W})}{\mathrm{n}(\mathrm{S})}=\frac{48}{100}\)
By Bayes’ theorem, the probability that a women selected at random is a graduate from the same town, is given by

Question 4.
If E₁ and E₂ are equally likely, mutually exclusive and exhaustive events and P(A/E₁) = 0.2, P(A/E₂) = 0.3. Find P(E₁/A).
Solution:
E₁ and E₂ are equally likely, mutually exclusive and exhaustive events.
∴ P(E₁) = P(E₂) = \(\frac{1}{2}\)

Question 5.
Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II?
Solution:
Let event J₁: Ball drawn from jar I,
event J₂: Ball drawn from jar II.
P(J₁) = P(head) = \(\frac{1}{2}\)
P(J₂) = P(tail) = \(\frac{1}{2}\)
Let event W: Ball drawn is white.
In Jar I, there are total 12 balls, out of which 5 balls are white.
∴ Probability that the ball drawn is white under the condtion that it is drawn from Jar I.
P(W/J₁) = \(\frac{{ }^{5} C_{1}}{{ }^{12} C_{1}}=\frac{5}{12}\)
Similarly, P(W/J₂) = \(\frac{{ }^{3} C_{1}}{{ }^{15} C_{1}}=\frac{3}{15}=\frac{1}{5}\)
Required probability = P(J₂/W)
By Bayes’ theorem,

Question 6.
A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability 0.10 of giving a (false) positive result when applied to a non-sufferer. It is estimated that 0.5% of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the probability that:
(i) given a positive result, the person is a sufferer.
(ii) given a negative result, the person is a non-sufferer.
Solution:
Let event T: Test positive
event S: Sufferer
P(S) = \(\frac{0.5}{100}\) = 0.005
∴ P(S’) = 1 – P(S) = 1 – 0.005 = 0.995
Since a probability of getting a positive result when applied to a person suffering from a disease is 0.95 and probability of getting positive result when applied to a non sufferer is 0.10.
∴ P(T/S) = 0.95 and P(T/S’) = 0.10
∴ P(T) = P(S) P(T/S) + P(S’) P(T/S’)
= 0.005 × 0.95 + 0.995 × 0.10
= 0.10425
∴ P(T’) = 1 – P(T) = 1 – 0.10425 = 0.8958
(i) Required probability = P(S/T)
By Bayes’ theorem,

(ii) P(T’/S’) = 1 – 0.1 = 0.9
Required probability = P(S’/T’)
By Bayes’ theorem

Question 7.
A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is 0.95. However occasionally children with flu also develop rash, whose chance are 0.08. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?
Solution:
Let the total number of sick children be 100.
∴ n(S) = 100.
Let event A: The child is sick with flu,
event B: The child is sick with measles,
event C: The child is sick with rash.
∴ n(A) = 80 and n(B) = 20
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{80}{100}=\frac{4}{5}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{100}=\frac{1}{5}\)
Since the chances of having rashes, if the child is suffering from measles is 0.95 and the chances of having rashes if the child has flu is 0.08,
P(C/B) = 0.95 = \(\frac{95}{100}\) and
P(C/A) = 0.08 = \(\frac{8}{100}\)
Required probability = P(B/C)
By Bayes’ theorem,

Question 8.
2% of the population have a certain blood disease of a serious form: 10% have it in a mild form; and 88% don’t have it at all. A new blood test is developed; the probability of testing positive is \(\frac{9}{10}\) if the subject has the
serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?
Solution:
Let event A₁: Disease in serious form,
event A₂: Disease in mild form,
event A₃: Subject does not have disease,
event B: Subject tests positive.
P(A₁) = 0.02, P(A₂) = 0.1, P(A₃) = 0.88
The probability of testing positive is \(\frac{9}{10}\) if the subject has the serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease.
∴ P(B/A₁) = 0.9, P(B/A₂) = 0.6, P(B/A₃) = 0.1
P(B) = P(A₁) P(B/A₁) + P(A₂) P(B/A₂) + P(A₃) P(B/A₃)
= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1
= 0.166
Required probability = P(A₁/B)
By Baye’s theorem

Question 9.
A box contains three coins: two fair coins and one fake two-headed coin. A coin is picked randomly from the box and tossed.
(i) What is the probability that it lands head up?
(ii) If happens to be head, what is the probability that it is the two-headed coin?
Solution:
Let event A: Fair coin is tossed,
event B: Fake coin is tossed
and event H: Head occur.
Clearly, a fair coin has one head.
∴ Probability that head occur under the condition that the fair coin is tossed = P(H/A) = \(\frac{1}{2}\)
Fake coin has two heads.
∴ Probability that head occur under the condition that the fake coin is tossed = P(H/B) = 1
n(A) = 2, n(B) = 1, n(S) = 3
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{3}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{3}\)
(i) Required probability
P(H) = P(A) P(H/A) + P(B) P(H/B)
= \(\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times 1\)
= \(\frac{1}{3}+\frac{1}{3}\)
= \(\frac{2}{3}\)

(ii) Required probability = P(B/H)
By Baye’s theorem

Question 10.
There are three social media groups on a mobile: Group I, Group II and Group III. The probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively. The probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively. Randomly one of the messages is opened and found a message on sports. What is the probability that the message was from Group III.
Solution:
Let event A: Message sent on sports by group I,
event B: Message sent on sports by group II,
event C: Message sent on sports by group III,
event E: Message is opened.
Given that the probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively and the probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
∴ P(A) = \(\frac{2}{5}\)
P(B) = \(\frac{1}{2}\)
P(C) = \(\frac{2}{3}\)
P(E/A) = \(\frac{1}{2}\)
P(E/B) = \(\frac{1}{4}\)
P(E/C) = \(\frac{1}{4}\)
Required probability = P(C/E)
By Baye’s theorem

Question 11.
Mr. X goes to office by Auto, Car and train. The probabilities of him travelling by these modes are \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{2}{7}\) respectively. The chances of him being late to the office are \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{4}\) respectively by Auto, Car and train. On one particular day he was late to the office. Find the probability that he travelled by car.
Solution:
Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively.
Let L be event that he is late.

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.3 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.3 Questions And Answers Maharashtra Board

Question 1.
A bag contains 3 red marbles and 4 blue marbles. Two marbles are drawn at random without replacement. If the first marble drawn is red, what is the probability that the second marble is blue?
Solution:
Total number of marbles = 3 + 4 = 7
Let event A: The first marble drawn is red.
∴ P(A) = \(\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{7} \mathrm{C}_{1}}=\frac{3}{7}\)
Let event B: The second marble drawn is blue.
Since the first red marble is not replaced in the bag, we now have 6 marbles out of which 4 are blue.
∴ Probability that the second marble is blue under the condition that the first red marble is not replaced in the bag = P(B/A) = \(\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{6} \mathrm{C}_{1}}=\frac{4}{6}=\frac{2}{3}\)
∴ Required probability = P(A ∩ B) = P(B/A) . P(A)
= \(\frac{2}{3} \times \frac{3}{7}\)
= \(\frac{2}{7}\)

Alternate Method:
Total number of marbles = 3 + 4 = 7
Two marbles are drawn at random without replacement.
∴ n(S) = \({ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}\) = 7 × 6 = 42
Let event A: The first marble is red and second marble is blue.
First red marble can be drawn from 3 red marbles in \({ }^{3} \mathrm{C}_{1}\) ways and second blue marble can be drawn from 4 blue marbles in \({ }^{4} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{3} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) = 3 × 4 = 12
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{42}=\frac{2}{7}\)

Question 2.
A box contains 5 green pencils and 7 yellow pencils. Two pencils are chosen at random from the box without replacement. What is the probability that both are yellow?
Solution:
Total number of pencils = 5 + 7 = 12
Let event A: The first pencil chosen is yellow.
∴ P(A) = \(\frac{{ }^{7} \mathrm{C}_{1}}{{ }^{12} \mathrm{C}_{1}}=\frac{7}{12}\)
Let event B: The second pencil chosen is yellow.
Since the first yellow pencil is not replaced in the box, we now have 11 pencils, out of which 6 are yellow.
∴ Probability that the second pencil is yellow under the condition that the first yellow pencil is not replaced in the box = P(B/A)
= \(\frac{{ }^{6} C_{1}}{{ }^{11} C_{1}}\)
= \(\frac{6}{11}\)
Required probability = P(A ∩ B)
= P(B/A) . P(A)
= \(\frac{6}{11} \times \frac{7}{12}\)
= \(\frac{7}{22}\)

Question 3.
In a sample of 40 vehicles, 18 are red, 6 are trucks, of which 2 are red. Suppose that a randomly selected vehicle is red. What is the probability it is a truck?
Solution:
One vehicle is selected from 40 vehicles.
Let event A: The selected vehicle is red.
There are total of 18 red vehicles.
∴ P(A) = \(\frac{{ }^{18} \mathrm{C}_{1}}{{ }^{40} \mathrm{C}_{1}}=\frac{18}{40}=\frac{9}{20}\)
Let event B: The selected vehicle is a truck.
There are total of 6 trucks.
Since 2 trucks are red, they are common between A and B.
∴ P(A ∩ B) = \(\frac{{ }^{2} \mathrm{C}_{1}}{{ }^{40} \mathrm{C}_{1}}=\frac{2}{40}=\frac{1}{20}\)
∴ Probability that the selected vehicle is a truck under the condition that it is red = P(B/A)
= \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\frac{1}{20}}{\frac{9}{20}}\)
= \(\frac{1}{9}\)

Question 4.
From a pack of well-shuffled cards, two cards are drawn at random. Find the probability that both the cards are diamonds when
(i) the first card drawn is kept aside.
(ii) the first card drawn is replaced in the pack.
Solution:
In a pack of 52 cards, there are 13 diamond cards.
Let event A: The first card drawn is a diamond card.
∴ P(A) = \(\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13}{52}=\frac{1}{4}\)
(i) Let event B: The second card drawn is a diamond card.
Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards.
Probability that the second card is a diamond card under the condition that the first diamond card is kept aside in the pack = P(B/A) = \(\frac{{ }^{12} \mathrm{C}_{1}}{{ }^{51} \mathrm{C}_{1}}=\frac{12}{51}=\frac{4}{17}\)
∴ Required probability = P(A ∩ B)
= P(B/A) . P(A)
= \(\frac{1}{4} \times \frac{4}{17}\)
= \(\frac{1}{17}\)

(ii) Let event B: The second card drawn is a diamond card.
Since the first diamond card is replaced in the pack, we now again have 52 cards, out of which 13 are diamond cards.
∴ Probability that the second card is a diamond card under the condition that the first diamond card is replaced in the pack = P(B/A) = \(\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13}{52}=\frac{1}{4}\)
Required probability = P(A ∩ B)
= P(B/A) . P(A)
= \(\frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{16}\)

Question 5.
A, B, and C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are \(\frac{3}{4}\), \(\frac{1}{2}\) and \(\frac{5}{8}\). Find the probability that the target
(a) is hit exactly by one of them.
(b) is not hit by any one of them.
(c) is hit.
(d) is exactly hit by two of them.
Solution:
Let event A: A can hit the target,
event B: B can hit the target,
event C: C can hit the target.

Since A, B, C are independent events,
A’, B’, C’ are also independent events.
(a) Let event W: Target is hit exactly by one of them.

(b) Let event X: Target is not hit by any one of them.
∴ P(X) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
= \(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}\)
= \(\frac{3}{64}\)

(c) Let event Y: Target is hit.
∴ P(Y) = 1 – P(target is not hit by any one of them)
= 1 – \(\frac{3}{64}\)
= \(\frac{61}{64}\)

(d) Let event Z: Target is hit by exactly two of them.

Question 6.
The probability that a student X solves a problem in dynamics is \(\frac{2}{5}\) and the probability that student Y solves the same problem is \(\frac{1}{4}\). What is the probability that
(i) the problem is not solved?
(ii) the problem is solved?
(iii) the problem is solved exactly by one of them?
Solution:
Let event A: Student X solves the problem in dynamics,
event B: Student Y solves the problem in dynamics.
∴ P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{4}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
P(B’) = 1 – P(B) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Since A and B are independent events,
A’ and B’ are also independent events.
(i) Let event C: Problem is not solved.
∴ P(C) = P(A’ ∩ B’)
= P(A’) . P(B’)
= \(\frac{3}{5} \times \frac{3}{4}\)
= \(\frac{9}{20}\)

(ii) Let event D: Problem is solved.
Problem can be solved if at least one of the two students solves the problem.
∴ P(D) = P(at least one student solves the problem)
= 1 – P(no student solves the problem)
= 1 – P(A’ ∩ B’)
= 1 – P(A’) P(B’)
= 1 – \(\frac{3}{5} \times \frac{3}{4}\)
= 1 – \(\frac{9}{20}\)
= \(\frac{11}{20}\)

(iii) Let event E: The problem is solved exactly by one of them.
∴ P(E) = P(A’ ∩ B) ∪ P(A ∩ B’)
= P(A’) . P(B) + P(A) . P(B’)
= \(\left(\frac{3}{5} \times \frac{1}{4}\right)+\left(\frac{2}{5} \times \frac{3}{4}\right)\)
= \(\frac{3}{20}+\frac{6}{20}\)
= \(\frac{9}{20}\)

Question 7.
A speaks truth in 80% of the cases and B speaks truth in 60% of the cases. Find the probability that they contradict each other in narrating an incident.
Solution:
Let event A : A speaks the truth,
event B : B speaks the truth.
∴ P(A) = \(\frac{80}{100}=\frac{4}{5}\)
and P(B) = \(\frac{60}{100}=\frac{3}{5}\)
P(A’) = 1 – P(A) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
and P(B’) = 1 – P(B) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
∴ P(A and B contradict each other) = P(A speaks the truth and B lies) + P (A lies and B speaks the truth)
= P(A ∩ B’) + P(A’ ∩ B)
= P(A) P(B’) + P(A’) P(B)
= \(\left(\frac{4}{5} \times \frac{2}{5}\right)+\left(\frac{1}{5} \times \frac{3}{5}\right)\)
= \(\frac{8}{25}+\frac{3}{25}\)
= \(\frac{11}{25}\)

Question 8.
Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery. The following table summarizes their response.

If one person from the 200 patients is selected at random, determine the probability
(a) that the person was satisfied given that the person had Throat surgery.
(b) that person was unsatisfied given that the person had eye surgery.
(c) the person had Throat surgery given that the person was unsatisfied.
Solution:
(a) Let event A: The patient was satisfied,
event B: The patient had throat surgery.
Given, n(S) = 200
n(A ∩ B) = 70
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{70}{200}\)
n(B) = 95
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{95}{200}\)
∴ Required probability = P(A / B)
= \(\frac{P(A \cap B)}{P(B)}\)
= \(\frac{\left(\frac{70}{200}\right)}{\left(\frac{95}{200}\right)}\)
= \(\frac{70}{95}\)
= \(\frac{14}{19}\)

Check:
Reduce the sample space to the set of throat patients only.
n(S) = 95
Let E : Patient had satisfactory throat surgery.
n(E) = 70
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{70}{95}=\frac{14}{19}\)

(b) Let event C : The patient was unsatisfied,
event D : The patient had a eye surgery.
Given, n(S) = 200
n(C ∩ D) = 15
∴ P(C ∩ D) = \(\frac{n(C \cap D)}{n(S)}=\frac{15}{200}\)
n(D) = 105
∴ P(D) = \(\frac{105}{200}\)
Required probability = P(C / D)
= \(\frac{P(C \cap D)}{P(D)}\)
= \(\frac{\left(\frac{15}{200}\right)}{\left(\frac{105}{200}\right)}\)
= \(\frac{1}{7}\)

(c) Let event F : The patient had a throat surgery,
event G : The patient was unsatisfied.
Given, n(S) = 200
n(F ∩ G) = 25
∴ P(F ∩ G) = \(\frac{n(F \cap G)}{n(S)}=\frac{25}{200}\)
n(G) = 40
∴ P(G) = \(\frac{n(G)}{n(S)}=\frac{40}{200}\)
∴ Required probability = P(F / G)
= \(\frac{P(F \cap G)}{P(G)}\)
= \(\frac{\left(\frac{25}{200}\right)}{\left(\frac{40}{200}\right)}\)
= \(\frac{5}{8}\)

Question 9.
Two dice are thrown together. Let A be the event ‘getting 6 on the first die’ and B be the event ‘getting 2 on the second die’. Are events A and B independent?
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: Getting 6 on the first die.
∴ A = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
Let event B : Gettting 2 on the second die.
∴ B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
∴ n(B) = 6
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
Now, A ∩ B = {(6, 2)}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{1}{36}\) …..(i)
P(A) × P(B) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\) ……..(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Question 10.
The probability that a man who is 45 years old will be alive till he becomes 70 is \(\frac{5}{12}\). The probability that his wife who is 40 years old will be alive till she becomes 65 is \(\frac{3}{8}\). What is the probability that, 25 years hence,
(a) the couple will be alive?
(b) exactly one of them will be alive?
(c) none of them will be alive?
(d) at least one of them will be alive?
Solution:
Let event A: The man will be alive till 70.
∴ P(A) = \(\frac{5}{12}\)
Let event B: The wife will be alive till 65.
∴ P(B) = \(\frac{3}{8}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{5}{12}\) = \(\frac{7}{12}\)
P(B’) = 1 – P(B) = 1 – \(\frac{3}{8}\) = \(\frac{5}{8}\)
Since A and B are independent events,
A’ and B’ are also independent events.
(a) Let event C : Both man and his wife will be alive.
∴ P(C) = P(A ∩ B) = P(A) . P(B)
= \(\frac{5}{12} \times \frac{3}{8}\)
= \(\frac{5}{32}\)

(b) Let event D: Exactly one of them will be alive.
∴ P(D) = P(A’ ∩ B) + P(A ∩ B’)
= P(A’) . P(B) + P(A) . P(B’)
= \(\left(\frac{7}{12} \times \frac{3}{8}\right)+\left(\frac{5}{12} \times \frac{5}{8}\right)\)
= \(\frac{21}{96}+\frac{25}{96}\)
= \(\frac{23}{48}\)

(c) Let event E: None of them will be alive.
∴ P(E) = P(A’ ∩ B’) = P(A’) . P(B’)
= \(\frac{7}{12} \times \frac{5}{8}\)
= \(\frac{35}{96}\)

(d) Let event F: At least one of them will be alive.
∴ P(F) = 1 – P(none of them will be alive)
= 1 – \(\frac{35}{96}\)
= \(\frac{61}{96}\)

Question 11.
A box contains 10 red balls and 15 green balls. Two balls are drawn in succession without replacement. What is the probability that,
(a) the first is red and the second is green?
(b) one is red and the other is green?
Solution:
Total number of balls = 10 + 15 = 25
(a) Let event A: First ball drawn is red.
∴ P(A) = \(\frac{{ }^{10} \mathrm{C}_{1}}{{ }^{25} \mathrm{C}_{1}}=\frac{10}{25}=\frac{2}{5}\)
Let event B: Second ball drawn is green.
Since the first red ball is not replaced in the box, we now have 24 balls, out of which 15 are green.
∴ Probability that the second ball is green under the condition that the first red ball is not replaced in the box = P(B/A) = \(\frac{{ }^{15} \mathrm{C}_{1}}{{ }^{24} \mathrm{C}_{1}}=\frac{15}{24}=\frac{5}{8}\)
∴ Required probability = P(A ∩ B) = P(B/A) . P(A)
= \(\frac{2}{5} \times \frac{5}{8}\)
= \(\frac{1}{4}\)

(b) To find the probability that one ball is red and the other is green, there are two possibilities:
First ball is red and second ball is green.
OR
The first ball is the green and the second ball is red.
From above, we get
P(First ball is red and second ball is green) = \(\frac{1}{4}\)
Similarly,
P(First ball is green and second ball is red) = \(\frac{{ }^{15} \mathrm{C}_{1}}{{ }^{25} \mathrm{C}_{1}} \times \frac{{ }^{10} \mathrm{C}_{1}}{{ }^{24} \mathrm{C}_{1}}=\frac{15}{25} \times \frac{10}{24}=\frac{1}{4}\)
∴ Required probability = P(First ball is red and second ball is green) + P(First ball is green and second ball is red)
= \(\frac{1}{4}\) + \(\frac{1}{4}\)
= \(\frac{1}{2}\)

Question 12.
A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that,
(a) both the balls are of the same colour?
(b) the balls are of a different colours?
Solution:
(a) Let event A: A yellow ball is drawn from each bag.
Probability of drawing one yellow ball from total 8 balls of first bag and that of drawing one yellow ball out of total 10 balls of second bag is
P(A) = \(\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{8} \mathrm{C}_{1}} \times \frac{{ }^{4} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}\) = \(\frac{3}{8} \times \frac{4}{10}=\frac{3}{20}\)
Let event B: A brown ball is drawn from each bag.
Probability of drawing one brown ball out of total 8 balls of first bag and that of drawing one brown ball out of total 10 balls of second bag is
P(B) = \(\frac{{ }^{5} \mathrm{C}_{1}}{{ }^{8} \mathrm{C}_{1}} \times \frac{{ }^{6} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}\) = \(\frac{5}{8} \times \frac{6}{10}=\frac{3}{8}\)
Since both the events are mutually exclusive events,
P(A ∩ B) = 0
∴ P(both the balls are of the same colour) = P(both are of yellow colour) or P(both are of brown colour)
= P(A) + P(B)
= \(\frac{3}{20}+\frac{3}{8}\)
= \(\frac{21}{40}\)

(b) P(both the balls are of different colour) = 1 – P(both the balls are of the same colour)
= 1 – \(\frac{21}{40}\)
= \(\frac{19}{40}\)

Question 13.
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement. What is the probability that at least one of them is black?
Solution:
Total number of balls in the um = 4 + 5 + 6 = 15
Two balls are drawn from 15 balls without replacement.
∴ n(S) = \({ }^{15} \mathrm{C}_{1} \times{ }^{14} \mathrm{C}_{1}\) = 15 × 14 = 210
Let event A: At least one ball is black.
i.e., the first ball is black, and the second ball is non-black or the first ball is non-black and the second ball is black, or both the first and second balls are black.
∴ n(A) = \({ }^{4} \mathrm{C}_{1} \times{ }^{11} \mathrm{C}_{1}+{ }^{11} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{1}\)
= 4 × 11 + 11 × 4 + 4 × 3
= 100
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{100}{210}=\frac{10}{21}\)

Check:
Required probability = 1 – P(no black ball in two balls)
= 1 – \(\frac{{ }^{11} C_{2}}{{ }^{15} C_{2}}=1-\frac{11 \times 10}{15 \times 14}=1-\frac{11}{21}=\frac{10}{21}\)

Question 14.
Three fair coins are tossed. What is the probability of getting three heads given that at least two coins show heads?
Solution:
When three fair coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ n(S) = 8
Let event A: Getting three heads.
∴ A = {HHH}
Let event B: Getting at least two heads.
∴ B = {HHT, HTH, THH, HHH}
∴ n(B) = 4
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{8}\)
Now, A ∩ B = {HHH}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{1}{8}\)
∴ Probability of getting three heads, given that at least two coins show heads, is given by
P(A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{\frac{1}{8}}{\frac{4}{8}}\)
= \(\frac{1}{4}\)

Question 15.
Two cards are drawn one after the other from a pack of 52 cards without replacement. What is the probability that both the cards are drawn are face cards?
Solution:
In a pack of52 cards, there are 12 face cards.
Let event A: The first card drawn is a face card.
∴ P(A) = \(\frac{{ }^{12} C_{1}}{{ }^{52} C_{1}}=\frac{12}{52}=\frac{3}{13}\)
Let event B: The second card drawn is a face card.
Since the first card is not replaced in the pack, we now have 51 cards, out of which 11 are face cards.
∴ Probability that the second card is a face card under the condition that the first card is not replaced in the pack = P(B/A) = \(\frac{{ }^{11} C_{1}}{{ }^{51} C_{1}}=\frac{11}{51}\)
∴ Required probability = P(A ∩ B) = P(B/A) . P(A)
= \(\frac{11}{51} \times \frac{3}{13}\)
= \(\frac{11}{221}\)

Question 16.
Bag A contains 3 red and 2 white balls and bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls are drawn are of the same colour.
Solution:
Let event C₁: The first ball drawn is red and from bag A,
event D₁: The first ball drawn is white and from bag A,
event E₁: The first ball drawn is red and from bag B,
event F₁: The first ball drawn is white and from bag B,
event C₂: Second ball drawn is red and from bag B,
event D₂: Second ball drawn is white and from bag B,
event E₂: Second ball drawn is red and from bag A,
event F₂: Second ball drawn is white and from bag A,
event G: Selecting bag A in the first place,
event H: Selecting bag B in the first place.
P(G) = P(H) = \(\frac{1}{2}\)
Let event X: Both the balls drawn are of same colour.
∴ P(X) = P(G) × P (X/G) + P(H) × P(X/H) …….(i)
If bag A is selected in first place, then In bag A, we have 5 balls, out of which 3 are red.
Probability of getting first red ball from bag A = P(C1) = \(\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{5} \mathrm{C}_{1}}=\frac{3}{5}\)
Since first red ball is put into the bag B, we now have 8 balls in bag B, out of which 3 are red.
∴ Probability of getting second red ball from bag B.
P(C₂/C₁) = \(\frac{{ }^{3} C_{1}}{{ }^{8} C_{1}}=\frac{3}{8}\)
Similarly, probability of getting first white ball from bag A = P(D₁) = \(\frac{{ }^{2} C_{1}}{{ }^{5} C_{1}}=\frac{2}{5}\)
and probability of getting second white ball form bag B = P(D₂/D₁) = \(\frac{{ }^{6} C_{1}}{{ }^{8} C_{1}}=\frac{6}{8}\)
∴ P(X/G) = P(C₁) P(C₂/C₁) + P(D₁) P(D₂/D₁)
= \(\frac{3}{5} \times \frac{3}{8}+\frac{2}{5} \times \frac{6}{8}\)
= \(\frac{21}{40}\) …..(ii)
Similarly, P(X/H) = P(E₁) P(E₂/E₁) + P(F₁) P(F₂/F₁)
= \(\frac{2}{7} \times \frac{4}{6}+\frac{5}{7} \times \frac{3}{6}\)
= \(\frac{23}{42}\) ………(iii)
From (i), (ii), (iii),
Required probability = \(\frac{1}{2} \times \frac{21}{40}+\frac{1}{2} \times \frac{23}{42}\)
= \(\frac{3604}{6720}\)
= \(\frac{901}{1680}\)

Question 17.
Activity: A bag contains 3 red and 5 white balls. Two balls are drawn at random one after the other without replacement. Find the probability that both the balls are white.
Solution:
Let, event A: The first ball drawn is white
event B: Second ball drawn is white.
P(A) = \(\frac{5}{8}\)
After drawing the first ball, without replacing it into the bag a second ball is drawn from the remaining 7 balls.
∴ P(B/A) = \(\frac{4}{7}\)
∴ P(Both balls are white) = P(A ∩ B)
= P(A) . P(B/A)
= \(\frac{5}{8}\) × \(\frac{4}{7}\)
= \(\frac{5}{14}\)

Question 18.
A family has two children. Find the probability that both the children are girls, given that at least one of them is a girl.
Solution:
A family has two children.
∴ Sample space S = {BB, BG, GB, GG}
∴ n(S) = 4
Let event A: At least one of the children is a girl.
∴ A = {GG, GB, BG}
∴ n(A) = 3
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{4}\)
Let event B: Both children are girls.
∴ B = {GG}
∴ n(B) = 1
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{4}\)
Also, A ∩ B = B
∴ P(A ∩ B) = P(B) = \(\frac{1}{4}\)
∴ Required probability = P(B/A)
= \(\frac{P(B \cap A)}{P(A)}\)
= \(\frac{\frac{1}{4}}{\frac{3}{4}}\)
= \(\frac{1}{3}\)

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.2 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.2 Questions And Answers Maharashtra Board

Question 1.
First, 6 faced die which is numbered 1 to 6 is thrown, then a 5 faced die which is numbered 1 to 5 is thrown. What is the probability that sum of the numbers on the upper faces of the dice is divisible by 2 or 3?
Solution:
When a 6 faced die and a 5 faced die are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(S) = 30
Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.
A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}
∴ n(A) = 15
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{15}{30}\)
Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.
B = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3)}
∴ n(B) = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{30}\)
Now,
A ∩ B = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{30}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{30}+\frac{10}{30}-\frac{5}{30}\)
= \(\frac{20}{30}\)
= \(\frac{2}{3}\)

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either black or a face card?
Solution:
One card can be drawn from the pack of 52 cards in \({ }^{52} \mathrm{C}_{1}\) = 52 ways.
∴ n(S) = 52
The pack of 52 cards consists of 26 red and 26 black cards.
(i) Let event A: A red card is drawn.
∴ Red card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26ways
∴ n(A) = 26
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{26}{52}\)
Let event B: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{26}{52}\)
Since A and B are mutually exclusive events,
P(A ∩ B) = 0
∴ Required probability
P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= 1

(ii) Let event A: A black card is drawn.
∴ Black card can be drawn in \({ }^{26} \mathrm{C}_{1}\) = 26 ways.
n(A) = 26
n(A) 26 n(S) ~ 52
Let event B: A face card is drawn.
There are 12 face cards in the pack of 52 cards.
∴ 1 face card can be drawn in \({ }^{12} \mathrm{C}_{1}\) = 12 ways.
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 black face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{6}{52}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\)
= \(\frac{32}{52}\)
= \(\frac{8}{13}\)

Question 3.
A girl is preparing for National Level Entrance exam and State Level Entrance exam for professional courses. The chances of her cracking National Level exam is 0.42 and that of State Level exam is 0.54. The probability that she clears both the exams is 0.11. Find the probability that
(i) she cracks at least one of the two exams.
(ii) she cracks only one of the two.
(iii) she cracks none.
Solution:
Let event A: The girl cracks the National Level exam.
∴ P(A) = 0.42
Let event B: The girl cracks the State Level exam.
∴ P(B) = 0.54
Also, P(A ∩ B) = 0.11
(i) P(the girl cracks at least one of the two exams)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.54 – 0.11
= 0.85

(ii) P(the girl cracks only one of the two exams)
= P(A) – P(B) – 2P(A ∩ B)
= 0.42 + 0.54 – 2(0.11)
= 0.74

(iii) P(the girl cracks none of the exams)
= P(A’ ∩ B’)
= P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – 0.85
= 0.15

Question 4.
A bag contains 75 tickets numbered from 1 to 75. One ticket is drawn at random. Find the probability that,
(i) number on the ticket is a perfect square or divisible by 4.
(ii) number on the ticket is a prime number or greater than 40.
Solution:
Out of the 75 tickets, one ticket can be drawn in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75
(i) Let event A: The number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(A) = 8
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{75}\)
Let event B: The number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72}
∴ n(B) = 18
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{18}{75}\)
Now, A ∩ B = {4, 16, 36, 64}
∴ n(A ∩ B) = 4
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{4}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{8}{75}+\frac{18}{75}-\frac{4}{75}\)
= \(\frac{22}{75}\)

(ii) Let event A: The number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A) = 21
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{21}{75}\)
Let event B: The number is greater than 40.
∴ B = {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75}
∴ n(B) = 35
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{35}{75}\)
Now,
A ∩ B = {41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(A ∩ B) = 9
∴ n(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{9}{75}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{21}{75}+\frac{35}{75}-\frac{9}{75}\)
= \(\frac{47}{75}\)

Question 5.
The probability that a student will pass in French is 0.64, will pass in Sociology is 0.45 and will pass in both is 0.40. What is the probability that the student will pass in at least one of the two subjects?
Solution:
Let event A: The student will pass in French.
∴ P(A) = 0.64
Let event B: The student will pass in Sociology.
∴ P(B) = 0.45
Also, P(A ∩ B) = 0.40
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.64 + 0.45 – 0.40
= 0.69

Question 6.
Two fair dice are thrown. Find the probability that the number on the upper face of the first die is 3 or sum of the numbers on their upper faces is 6.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The number on the upper face of the first die is 3.
∴ A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{36}\)
Let event B: Sum of the numbers on their upper faces is 6.
∴ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
∴ n(B) = 5
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{36}\)
Now, A ∩ B = {(3, 3)}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{1}{36}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{36}+\frac{5}{36}-\frac{1}{36}\)
= \(\frac{10}{36}\)
= \(\frac{5}{18}\)

Question 7.
For two events A and B of a sample space S, if P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\) and P(A ∪ B) = \(\frac{5}{8}\). Find the value of the following.
(a) P(A ∩ B)
(b) P(A’ ∩ B’)
(c) P(A’ ∪ B’)
Solution:
Here, P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{5}{8}\)
(a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{3}{8}+\frac{1}{2}-\frac{5}{8}\)
= \(\frac{1}{4}\)

(b) P(A’ ∩ B’) = P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

(c) P(A’ ∪ B’) = P(A ∩ B)’
= 1 – P(A ∩ B)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Question 8.
For two events A and B of a sample space S, if P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\) and P(B’) = \(\frac{1}{3}\), then find P(A).
Solution:
Here, P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\), P(B’) = \(\frac{1}{3}\)
P(B) = 1 – P(B’)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(\frac{5}{6}\) = P(A) + \(\frac{2}{3}-\frac{1}{3}\)
∴ \(\frac{5}{6}\) = P(A) + \(\frac{1}{3}\)
∴ P(A) = \(\frac{5}{6}-\frac{1}{3}\) = \(\frac{1}{2}\)

Question 9.
A bag contains 5 red, 4 blue and an unknown number m of green balls. If the probability of getting both the balls green, when two balls are selected at random is \(\frac{1}{7}\), find m.
Solution:
Total number of balls in the bag = 5 + 4 + m = 9 + m
Two balls are selected from (9 + m) balls in \({ }^{9+m} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{9+m} \mathrm{C}_{2}\)
Let event A: The two balls selected are green.
∴ 2 balls can be selected from m balls in \({ }^{\mathrm{m}} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{\mathrm{m}} \mathrm{C}_{2}\)

(9 + m)(8 + m) = 7m(m – 1)
72 + 9m + 8m + m2 = 7m2 – 7m
6m2 – 24m – 72 = 0
m2 – 4m – 12 = 0
(m – 6)(m + 2) = 0
m = 6 or m = -2
Since number of balls cannot be negative, m ≠ -2
∴ m = 6

Question 10.
From a group of 4 men, 4 women and 3 children, 4 persons are selected at random. Find the probability that,
(i) no child is selected.
(ii) exactly 2 men are selected.
Solution:
The group consists of 4 men, 4 women and 3 children, i.e., 4 + 4 + 3 = 11 persons.
4 persons are to be selected from this group.
∴ 4 persons can be selected from 11 persons in \({ }^{11} \mathrm{C}_{4}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{4}\)
(i) Let event A: No child is selected.
∴ 4 persons can be selected from 4 men and 4 women, i.e., from 8 persons in \({ }^{8} \mathrm{C}_{4}\) ways.
∴ n(A) = \({ }^{8} \mathrm{C}_{4}\)

(ii) Let event B: Exactly 2 men are selected.
∴ 2 men are selected from 4 men in \({ }^{4} \mathrm{C}_{2}\) ways, and remaining 2 persons are selected from 7 persons (i.e., 4 women and 3 children) in \({ }^{7} \mathrm{C}_{2}\) ways.

Question 11.
A number is drawn at random from the numbers 1 to 50. Find the probability that it is divisible by 2 or 3 or 10.
Solution:
One number can be drawn at random from the numbers 1 to 50 in \({ }^{50} \mathrm{C}_{1}\) = 50 ways.
∴ n(S) = 50
Let event A: The number drawn is divisible by 2.
∴ A = {2, 4, 6, 8, 10, …, 48, 50}
∴ n(A) = 25
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{25}{50}\)
Let event B: The number drawn is divisible by 3.
B = {3, 6, 9, 12, …, 48}
∴ n(B) = 16
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{16}{50}\)
Let event C: The number drawn is divisible by 10.
C = {10, 20, 30, 40, 50}
∴ n(C) = 5
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{5}{50}\)
Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}
∴ n(A ∩ B) = 8
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{8}{50}\)
B ∩ C = {30}
∴ n(B ∩ C) = 1
∴ P(B ∩ C) = \(\frac{\mathrm{n}(\mathrm{B} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{1}{50}\)
A ∩ C = {10, 20, 30, 40, 50}
∴ n(A ∩ C) = 5
∴ P(A ∩ C) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
A ∩ B ∩ C = {30}
∴ n(A ∩ B ∩ C) = 1
∴ P(A ∩ B ∩C) = \(\frac{n(A \cap B \cap C)}{n(S)}=\frac{1}{50}\)
∴ P(the number is divisible by 2 or 3 or 10)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50}\)
= \(\frac{33}{50}\)

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.1 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.1 Questions And Answers Maharashtra Board

Question 1.
There are four pens: Red, Green, Blue, and Purple in a desk drawer of which two pens are selected at random one after the other with replacement. State the sample space and the following events.
(a) A : Select at least one red pen.
(b) B : Two pens of the same colour are not selected.
Solution:
The drawer contains 4 pens out of which one is red (R), one is green (G), one is blue (B) and the other one is purple (P).
From this drawer, two pens are selected one after the other with replacement.
∴ The sample space S is given by
S = {RR, RG, RB, RP, GR, GG, GB, GP, BR, BG, BB, BP, PR, PG, PB, PP}
(a) A : Select at least one red pen.
At least one means one or more than one.
∴ A = {RR, RG, RB, RP, GR, BR, PR}

(b) B : Two pens of the same colour are not selected.
B = {RG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, PB}

Question 2.
A coin and a die are tossed simultaneously. Enumerate the sample space and the following events.
(a) A : Getting a tail and an odd number.
(b) B : Getting a prime number.
(c) C : Getting head and a perfect square.
Solution:
When a coin and a die are tossed simultaneously, the sample space S is given by
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
(a) A : Getting a tail and an odd number.
∴ A = {(T, 1), (T, 3), (T, 5)}

(b) B : Getting a prime number.
∴ B = {(H, 2), (H, 3), (H, 5), (T, 2), (T, 3), (T, 5)}

(c) C : Getting a head and a perfect square.
∴ C = {(H, 1), (H, 4)}

Question 3.
Find n(S) for each of the following random experiments.
(a) From an urn containing 5 gold and 3 silver coins, 3 coins are drawn at random.
(b) 5 letters are to be placed into 5 envelopes such that no envelope is empty.
(c) 6 books of different subjects are arranged on a shelf.
(d) 3 tickets are drawn from a box containing 20 lottery tickets.
Solution:
(a) There are 5 gold and 3 silver coins, i.e., 8 coins.
3 coins can be drawn from these 8 coins in \({ }^{8} \mathrm{C}_{3}\) ways.
∴ n(s) = \({ }^{8} \mathrm{C}_{3}=\frac{8 !}{5 ! 3 !}=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}=56\)

(b) 5 letters have to be placed in 5 envelopes in such a way that no envelope is empty.
∴ The first letter can be placed into 5 envelopes in 5 different ways, the second letter in 4 ways.
Similarly, the third, fourth and fifth letters can be placed in 3 ways, 2 ways and 1 way, respectively.
∴ Total number of ways = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(c) 6 books can be arranged on a shelf in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(d) 3 tickets are drawn at random from 20 tickets.
∴ 3 tickets can be selected in \({ }^{20} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{3}=\frac{20 !}{17 ! 3 !}=\frac{20 \times 19 \times 18 \times 17 !}{17 ! \times 3 \times 2 \times 1}\) = 1140

Question 4.
Two fair dice are thrown. State the sample space and write the favourable outcomes for the following events.
(a) A : Sum of numbers on two dice is divisible by 3 or 4.
(b) B : The sum of numbers on two dice is 7.
(c) C : Odd number on the first die.
(d) D : Even number on the first die.
(e) Check whether events A and B are mutually exclusive and exhaustive.
(f) Check whether events C and D are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) A: Sum of the numbers on two dice is divisible by 3 or 4.
∴ A = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(b) B: Sum of the numbers on two dice is 7.
∴ B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(c) C: Odd number on the first die.
∴ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(d) D: Even number on the first die.
∴ D = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(e) A and B are mutually exclusive events as A ∩ B = Φ.
A ∪ B = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ A and B are not exhaustive events as A ∪ B ≠ S.

(f) C and D are mutually exclusive events as C ∩ D = Φ.
C ∪ D = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
∴ C and D are exhaustive events.

Question 5.
A bag contains four cards marked as 5, 6, 7, and 8. Find the sample space if two cards are drawn at random
(a) with replacement.
(b) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8. Two cards are to be drawn from this bag.
(a) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6, 8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(b) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

Question 6.
A fair die is thrown two times. Find the probability that
(a) the sum of the numbers on them is 5.
(b) the sum of the numbers on them is at least 8.
(c) the first throw gives a multiple of 2 and the second throw gives a multiple of 3.
(d) product of numbers on them is 12.
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}
∴ n(S) = 36
(a) Let event A: Sum of the numbers on uppermost face is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(b) Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8)
∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(c) Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3.
∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(d) Let event D: The product of the numbers on uppermost face is 12.
∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(D) = 4
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

Question 7.
Two cards are drawn from a pack of 52 cards. Find the probability that
(a) one is a face card and the other is an ace card.
(b) one is a club and the other is a diamond.
(c) both are from the same suit.
(d) both are red cards.
(e) one is a heart card and the other is a non-heart card.
Solution:
Two cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{2}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{2}\)

(a) Let event A: Out of the two cards drawn, one is a face card and the other is an ace card.
There are 12 face cards and 4 ace cards in a pack of 52 cards.
∴ One face card can be drawn from 12 face cards in \({ }^{12} \mathrm{C}_{1}\) ways and one ace card can be drawn from 4 ace cards in \({ }^{4} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\)

(b) Let event B: Out of the two cards drawn, one is club and the other is a diamond card.
There are 13 club cards and 13 diamond cards.
∴ One club card can be drawn from 13 club cards in \({ }^{13} \mathrm{C}_{1}\) ways and one diamond card can be drawn from 13 diamond cards in \({ }^{13} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{{ }^{13} C_{1} \times{ }^{13} C_{1}}{{ }^{52} C_{2}}=\frac{13 \times 13}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{102}\)

(c) Let event C: Both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 2 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{2}\) ways.
∴ n(C) = \({ }^{13} \mathrm{C}_{2} \times 4\)
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 13 \times 12}{52 \times 51}=\frac{4}{17}\)

(d) Let event D: Both the cards drawn are red.
There are 26 red cards in the pack of 52 cards.
∴ 2 cards can be drawn from them in \({ }^{26} \mathrm{C}_{2}\) ways.
∴ n(D) = \({ }^{26} \mathrm{C}_{2}\)
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{{ }^{26} C_{2}}{{ }^{52} C_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

(e) Let event E: Out of the two cards drawn, one is heart and other is non-heart.
There are 13 heart cards in a pack of 52 cards, i.e., 39 cards are non-heart.
∴ One heart card can be drawn from 13 hdart cards in \({ }^{13} \mathrm{C}_{1}\) ways and one non-heart card can be drawn from 39 cards in \({ }^{39} \mathrm{C}_{1}\) ways.
∴ n(E) = \({ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}\)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}=\frac{13 \times 39}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{34}\)

Question 8.
Three cards are drawn from a pack of 52 cards. Find the chance that
(a) two are queen cards and one is an ace card.
(b) at least one is a diamond card.
(c) all are from the same suit.
(d) they are a king, a queen, and a jack.
Solution:
3 cards can be drawn from a pack of 52 cards in \({ }^{52} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{52} \mathrm{C}_{3}\)

(a) Let event A: Out of the three cards drawn, 2 are queens and 1 is an ace card.
There are 4 queens and 4 aces in a pack of 52 cards.
∴ 2 queens can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{2}\) ways and 1 ace can be drawn out of 4 aces in \({ }^{4} \mathrm{C}_{1}\) ways.

(b) Let event B: Out of the three cards drawn, at least one is a diamond.
∴ B’ is the event that all 3 cards drawn are non-diamond cards.
In a pack of 52 cards, there are 39 non-diamond cards.
∴ 3 non-diamond cards can be drawn in \({ }^{39} \mathrm{C}_{3}\) ways.

(c) Let event C: All the cards drawn are from the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ 3 cards can be drawn from the same suit in \({ }^{13} \mathrm{C}_{3}\) ways.

(d) Let event D: The cards drawn are a king, a queen, and a jack.
There are 4 kings, 4 queens and 4 jacks in a pack of 52 cards.
∴ 1 king can be drawn from 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways,
1 queen can be drawn from 4 queens in \({ }^{4} \mathrm{C}_{1}\) ways and
1 jack can be drawn from 4 jacks in \({ }^{4} \mathrm{C}_{1}\) ways.

Question 9.
From a bag containing 10 red, 4 blue, and 6 black balls, a ball is drawn at random. Find the probability of drawing
(a) a red bail.
(b) a blue or black ball.
(c) not a black ball.
Solution:
The bag contains 10 red, 4 blue, and 6 black balls,
i.e., 10 + 4 + 6 = 20 balls.
One ball can be drawn from 20 balls in \({ }^{20} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{20} \mathrm{C}_{1}\) = 20

(a) Let event A: Ball drawn is red.
There are total 10 red balls.
∴ 1 red ball can be drawn from 10 red balls in \({ }^{10} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{10} \mathrm{C}_{1}\) = 10
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(b) Let event B: The ball drawn is blue or black.
There are 4 blue and 6 black balls.
∴ 1 blue ball can be drawn from 4 blue balls in \({ }^{4} \mathrm{C}_{1}\) ways
or 1 black ball can be drawn from 6 black balls in \({ }^{6} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{4} \mathrm{C}_{1}\) + \({ }^{6} \mathrm{C}_{1}\) = 4 + 6 = 10
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{10}{20}=\frac{1}{2}\)

(c) Let event C: Ball drawn is not black,
i.e., ball drawn is red or blue.
There are total 14 red and blue balls.
∴ 1 ball can be drawn from 14 balls in \({ }^{14} \mathrm{C}_{1}\) ways.
∴ n(C) = \({ }^{14} \mathrm{C}_{1}\) = 14
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{14}{20}=\frac{7}{10}\)

Question 10.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. Find the probability that,
(a) number on the ticket is divisible by 6.
(b) the number on the ticket is a perfect square.
(c) the number on the ticket is prime.
(d) the number on the ticket is divisible by 3 and 5.
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in \({ }^{75} \mathrm{C}_{1}\) = 75 ways.
∴ n(S) = 75

(a) Let event A: Number on the ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{75}=\frac{4}{25}\)

(b) Let event B: Number on the ticket is a perfect square.
∴ B = (1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{75}\)

(c) Let event C: Number on the ticket is a prime number.
∴C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}

(d) Let event D: Number on the ticket is divisible by 3 and 5,
i.e., divisible by L.C.M. of 3 and 5,
i.e., 15.
∴D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{5}{75}=\frac{1}{15}\)

Question 11.
What is the chance that a leap year, selected at random, will contain 53 Sundays?
Solution:
A leap year consists of 366 days.
It has 52 complete weeks and two more days.
These two days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)}.
∴ n(S) = 7
Let event E : There are 53 Sundays.
∴ E = {(Sun, Mon), (Sat, Sun)}
∴ n(E) = 2
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{7}\)

Question 12.
Find the probability of getting both red balls, when from a bag containing 5 red and 4 black balls, two balls are drawn,
(i) with replacement
(ii) without replacement
Solution:
The bag contains 5 red and 4 black balls,
i.e., 5 + 4 = 9 balls.
(i) 2 balls can be drawn from 9 balls with replacement in \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{9} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{1}\) = 9 × 9 = 81
Let event A: Balls drawn are red.
2 red balls can be drawn from 5 red balls with replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) ways.
∴ n(A) = \({ }^{5} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1}\) = 5 × 5 = 25
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{25}{81}\)

(ii) 2 balls can be drawn from 9 balls without replacement in \({ }^{9} C_{1} \times{ }^{8} C_{1}\) ways.
∴ n(S) = \({ }^{9} C_{1} \times{ }^{8} C_{1}\) = 9 × 8 = 72
2 red balls can be drawn from 5 red balls without replacement in \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) ways.
∴ n(B) = \({ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}\) = 5 × 4 = 20
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{72}=\frac{5}{18}\)

Question 13.
A room has three sockets for lamps. From a collection of 10 bulbs of which 6 are defective. At night a person selects 3 bulbs, at random and puts them in sockets. What is the probability that
(i) room is still dark.
(ii) the room is lit.
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 bulbs in \({ }^{10} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{10} \mathrm{C}_{3}\)

(i) Let event A: The room is dark.
For event A to happen the bulbs should be selected from the 6 defective bulbs. This can be done in \({ }^{6} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{6} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{6} C_{3}}{{ }^{10} C_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}\)

(ii) Let event A’: The room is lit.
∴ P(Room is lit) = 1 – P(Room is not lit)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)

Question 14.
Letters of the word MOTHER are arranged at random. Find the probability that in the arrangement
(a) vowels are always together.
(b) vowels are never together.
(c) O is at the beginning and end with T.
(d) starting with a vowel and ending with a consonant.
Solution:
There are 6 letters in the word MOTHER.
These letters can be arranged among themselves in \({ }^{6} \mathrm{P}_{6}\) = 6! ways.
∴ n(S) = 6!
(a) Let event A: Vowels are always together.
The word MOTHER consists of 2 vowels (O, E) and 4 consonants (M, T, H, R).
2 vowels can be arranged among themselves in \({ }^{2} \mathbf{P}_{2}\) = 2! ways.
Let us consider 2 vowels as one group.
This one group with 4 consonants can be arranged in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
∴ n(A) = 2! × 5!
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2 ! \times 5 !}{6 !}=\frac{1}{3}\)

(b) Let event B: Vowels are never together.
4 consonants create 5 gaps, in which vowels are arranged.
Consider the following arrangement of consonants
_C_C_C_C_
2 vowels can be arranged in 5 gaps in \({ }^{5} \mathbf{P}_{2}\) ways.
Also 4 consonants can be arranged among themselves in \({ }^{4} \mathbf{P}_{4}\) = 4! ways.
∴ n(B) = 4! × \({ }^{5} \mathbf{P}_{2}\)
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4 ! \times{ }^{5} P_{2}}{6 !}=\frac{4 ! \times 5 \times 4}{6 \times 5 \times 4 !}=\frac{2}{3}\)

(iii) Let event C: Word begin with O and end with T.
Thus first and last letters can be arranged in one way each and the remaining 4 letters can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways
∴ n(C) = 4! × 1 × 1 = 4!
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{4 !}{6 !}=\frac{1}{30}\)

(d) Let event D: Word starts with a vowel and ends with a consonant.
There are 2 vowels and 4 consonants in the word MOTHER.
∴ The first place can be arranged in 2 different ways and the last place can be arranged in 4 different ways.
Now, the remaining 4 letters (3 consonants and 1 vowel) can be arranged in \({ }^{4} \mathrm{P}_{4}\) = 4! ways.
∴ n(D) = 2 × 4 × 4!
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{2 \times 4 \times 4 !}{6 !}=\frac{4}{15}\)

Question 15.
4 letters are to be posted in 4 post boxes. If any number of letters can be posted in any of the 4 post boxes, what is the probability that each box contains only one letter?
Solution:
There are 4 letters and 4 post boxes.
Since any number of letters can be posted in all 4 post boxes,
so each letter can be posted in different ways.
∴ n(S) = 4 × 4 × 4 × 4
Let event A: Each box contains only one letter.
∴ 1st letter can be posted in 4 different ways.
Since each box contains only one letter, 2nd letter can be posted in 3 different ways.
Similarly, 3rd and 4th letters can be posted in 2 different ways and 1 way respectively.
∴ n(A) = 4 × 3 × 2 × 1
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\)

Question 16.
15 professors have been invited for a round table conference by the Vice-chancellor of a university. What is the probability that two particular professors occupy the seats on either side of the Vice-chancellor during the conference?
Solution:
Since a Vice-chancellor invited 15 professors for a round table conference, there were all 16 persons in the conference.
These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways.
∴ n(S) = 15!
Let event A: Two particular professors be seated on either side of the Vice-chancellor.
Those two particular persons sit on either side of a Vice chancellor in \({ }^{2} \mathrm{P}_{2}\) = 2! ways.
Thus the remaining 13 persons can be arranged in \({ }^{13} \mathrm{P}_{13}\) = 13! ways.
∴ n(A) = 13! 2!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13 ! \times 2 !}{15 !}=\frac{13 ! \times 2 \times 1}{15 \times 14 \times 13 !}=\frac{1}{105}\)

Question 17.
A bag contains 7 black and 4 red balls. If 3 balls are drawn at random, find the probability that
(i) all are black.
(ii) one is black and two are red.
Solution:
The bag contains 7 black and 4 red balls,
i.e., 7 + 4 = 11 balls.
∴ 3 balls can be drawn out of 11 balls in \({ }^{11} \mathrm{C}_{3}\) ways.
∴ n(S) = \({ }^{11} \mathrm{C}_{3}\)
(i) Let event A: All 3 balls drawn are black.
There are 7 black balls.
∴ 3 black balls can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{3}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{3}\)
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{7} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{3}}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}\)

(ii) Let event B: Out of 3 balls drawn, one is black and two are red.
There are 7 black and 4 red balls.
∴ One black ball can be drawn from 7 black balls in \({ }^{7} \mathrm{C}_{1}\) ways and 2 red balls can be drawn from 4 red balls in \({ }^{4} \mathrm{C}_{2}\) ways.
∴ n(A) = \({ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}\)
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{7} C_{1} \times{ }^{4} C_{2}}{{ }^{11} C_{3}}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 10 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\)

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Miscellaneous Exercise 8 Questions and Answers.

11th Maths Part 1 Measures of Dispersion Miscellaneous Exercise 8 Questions And Answers Maharashtra Board

(I) Select the correct option from the given alternatives:

Question 1.
If there are 10 values each equal to 10, then S.D. of these values is _________
(A) 100
(B) 20
(C) 0
(D) 6
Answer:
(C) 0
Hint:
Var (X) = \(\sigma_{x}^{2}=\frac{\sum x_{i}^{2}}{\mathrm{n}}-(\bar{x})^{2}\)
= \(\frac{1000}{10}\) – 100
= 0
∴ S.D. = 0

Question 2.
The number of patients who visited cardiologists are 13, 17, 11, 15 in four days, then variance (approximately) is
(A) 5 patients
(B) 4 patients
(C) 10 patients
(D) 15 patients
Answer:
(A) 5 patients

Question 3.
If the observations of a variable X are, -4, -20, -30, -44 and -36, then the value of the range will be:
(A) -48
(B) 40
(C) -40
(D) 48
Answer:
(B) 40

Question 4.
The standard deviation of a distribution divided by the mean of the distribution and expressed in percentage is called:
(A) Coefficient of Standard deviation
(B) Coefficient of skewness
(C) Coefficient of quartile deviation
(D) Coefficient of variation
Answer:
(D) Coefficient of variation

Question 5.
If the S.D. of first n natural numbers is √2, then the value of n must be
(A) 5
(B) 4
(C) 7
(D) 6
Answer:
(A) 5

Question 6.
The positive square root of the mean of the squares of the deviations of observations from their mean is called:
(A) Variance
(B) Range
(C) S.D.
(D) C.V.
Answer:
(C) S.D.

Question 7.
The variance of 19, 21, 23, 25 and 27 is 8. The variance of 14, 16, 18, 20 and 22 is:
(A) Greater than 8
(B) 8
(C) Less than 8
(D) 8 – 5 = 3
Answer:
(B) 8

Question 8.
For any two numbers, SD is always
(A) Twice the range
(B) Half of the range
(C) Square of the range
(D) None of these
Answer:
(B) Half of the range

Question 9.
Given the heights (in cm) of two groups of students:
Group A: 131 cm, 150 cm, 147 cm, 138 cm, 144 cm
Group B: 139 cm, 148 cm, 132 cm, 151 cm, 140 cm
Which of the following is/are true?
I. The ranges of the heights of the two groups of students are the same.
II. The means of the heights of the two groups of students are the same.
(A) I only
(B) II only
(C) Both I and II
(D) None
Answer:
(C) Both I and II

Question 10.
The standard deviation of data is 12 and mean is 72, then the coefficient of variation is
(A) 13.67%
(B) 16.67%
(C) 14.67%
(D) 15.67%
Answer:
(B) 16.67%

(II) Answer the following:

Question 1.
Find the range for the following data.
76, 57, 80, 103, 61, 63, 89, 96, 105, 72
Solution:
Here, largest value (L) = 105, smallest value (S) = 57
∴ Range = L – S
= 105 – 57
= 48

Question 2.
Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Solution:
Here, largest value (L) = 203, smallest value (S) = 114
∴ Range = L – S
= 203 – 114
= 89

Question 3.
Given below is the frequency distribution of weekly wages of 400 workers. Find the range.

Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Question 4.
Find the range of the following data.

Solution:
Here, upper limit of the highest class (L) = 175
lower limit of the lowest class (S) = 115
∴ Range = L – S
= 175 – 115
= 60

Question 5.
Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 21, 25
Solution:

Question 6.
Find variance and S.D. for the following set of numbers.
125, 130, 150, 165, 190, 195, 210, 230, 245, 260
Solution:

Question 7.
Following data gives no. of goals scored by a team in 90 matches. Compute the standard deviation.

Solution:

Question 8.
Compute the variance and S.D. for the following data:

Solution:

Question 9.
Calculate S.D. from the following data.

Solution:
Since data is not continuous, we have to make it continuous.
let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-54.5}{10}\)
Calculation of variance of u:

Question 10.
Given below is the frequency distribution of marks obtained by 100 students. Compute arithmetic mean and S.D.

Solution:
Since data is not continuous, we have to make it continuous.
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-74.5}{10}\)
Calculation of variance of u:


∴ Var (X) = h² var (u)
= (10)² × 1.4875
= 100 × 1.4875
= 148.75
∴ S.D. = σ_x = √Var(X)
= √148.75
= 12.2

Question 11.
The arithmetic mean and standard deviation of a series of 20 items were calculated by a student as 20 cm and 5 cm respectively. But while calculating them, item 13 was misread as 30. Find the corrected mean and standard deviation.
Solution:
n = 20, \(\bar{x}\) = 20, σ_x = 5 …..(given)
\(\bar{x}=\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}\)
∴ \(\sum x_{\mathrm{i}}=\mathrm{n} \bar{x}\) = 20 × 20 = 400
Corrected Σx_i = Σx_i – (incorrect observation) + (correct observation)
= 400 – 30 + 13
= 383

Corrected \(\sum x_{\mathrm{i}}^{2}\) = \(\sum x_{\mathrm{i}}^{2}\) – (incorrect observation)² + (correct observation)²
= 8500 – (30)² + (13)²
= 8500 – 900 + 169
= 7769

Question 12.
The mean and S.D. of a group of 50 observations are 40 and 5 respectively. If two more observations 60 and 72 are added to the set, find the mean and S.D. of 52 items.
Solution:

Question 13.
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Solution:
Let n₁ and n₂ be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{c}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{2}\) = 62, σ₁ = 8, σ₂ = 10
Here, n₁ + n₂ = n
n₁ + n₂ = 200 ……(i)
Combined mean (\(\bar{x}_{c}\)) = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{i}+n_{2}}\)
65 = \(\frac{\mathrm{n}_{1}(70)+\mathrm{n}_{2}(62)}{200}\) …… [From (i)]
70n₁ + 62n₂ = 13000
35n₁ + 31n₂ = 6500 …..(ii)
Solving (i) and (ii), we get
n₁ = 75, n₂ = 125
Number of boys = 75

Question 14.
From the following data available for 5 pairs of observations of two variables x and y, obtain combined S.D. for all 10 observations.

Solution:

Question 15.
Calculate the coefficient of variation of the following data.
23, 27, 25, 28, 21, 14, 16, 12, 18, 16
Solution:

Question 16.
The following data relates to the distribution of weights of 100 boys and 80 girls in a school.

Which of the two is more variable?
Solution:

Here, C.V. of boys > C.V. of girls
∴ The Series of boys is more variable.

Question 17.
The mean and standard deviations of the two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.
Solution:

Here, C.V. (I) > C.V. (II)
∴ The brand I is more variable.

Question 18.
Calculate the coefficient of variation for the data given below:

Solution:

Question 19.
Calculate the coefficient of variation for the data given below:

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-6500}{1000}\)
Calculation of variance of u:

Question 20.
Compute the coefficient of variations for the following data to show whether the variation is greater in the field or in the area of the field.

Solution:


∴ the variation is greater in the area of the field.

Question 21.
There are two companies U and V which manufacture cars. A sample of 40 cars each from these companies is taken and the average running life (in years) is recorded.

Which company shows greater consistency?
Solution:
Let f₁ denote no. of cars of company U and f₂ denote no. of cars of company V.



Here, C.V. (U) < C.V. (V)
∴ Company U shows greater consistency in performance.

Question 22.
The means and S.D. of weights and heights of 100 students of a school are as follows.

Which shows more variability, weights, or heights?
Solution:

Here, C.V. for weight < C.V. for height
∴ Height shows more variability.

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.3 Questions and Answers.

11th Maths Part 1 Measures of Dispersion Exercise 8.3 Questions And Answers Maharashtra Board

Question 1.
The means of two samples of sizes 60 and 120 respectively are 35.4 and 30.9 and the standard deviations are 4 and 5. Obtain the standard deviation of the sample of size 180 obtained by combining the two samples.
Solution:

Question 2.
For certain data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are:
85, 91, 96, 88, 98, 82
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variance of 2.5%. What is the standard deviation of their heights?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%

∴ The standard deviation of students’ height is 3.76 cm.

Question 6.
Two workers on the same job show the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since C.V. (P) < C.V.(Q), Worker P is more consistent regarding the time required to complete the job. (ii) Since \(\overline{\mathrm{p}}>\overline{\mathrm{q}}\),
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since \(\bar{x}_{1}>\bar{x}_{2}\),
i.e., average weekly wages are more for the first department.
∴ The first department has a larger bill.

(ii) Since C.V. (1) < C.V. (2),
second department is less consistent.
∴ The second department has larger variability in wages.

Question 8.
The following table gives the weights of the students of two classes. Calculate the coefficient of variation of the two distributions. Which series is more variable?

Solution:
Let x denote the data of class A and y denote the data of class B.
Calculation of S.D. for class A:



Since C.V. (Y) > C.V.(X),
C.V. (B) > C.V. (A)
∴ Series B is more variable.

Question 9.
Compute the coefficient of variation for team A and team B.

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.




Since C.V. of team A > C.V. of team B,
Team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M),
The subject statistics show higher variability in marks.

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.2 Questions and Answers.

11th Maths Part 1 Measures of Dispersion Exercise 8.2 Questions And Answers Maharashtra Board

Question 1.
Find variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data:
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is

Calculation of variance and S.D.

Question 2.
Find variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Question 3.
Compute variance and standard deviation for the following data:

Solution:

Question 4.
Compute the variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-34}{1}\)
Calculation of variance of u:

Question 5.
Following data gives ages of 100 students in a college. Calculate variance and S.D.

Solution:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-19}{1}\)

Question 6.
Find mean, variance and S.D. of the following data.

Solution:


Alternate Method:
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-55}{10}\)
Calculation of variance of u:

Question 7.
Find the variance and S.D. of the following frequency distribution which gives the distribution of 200 plants according to their height.

Solution:
Since data is not continuous, we have to make it continuous.
Let u = \(\frac{x-\mathrm{A}}{\mathrm{h}}=\frac{x-31}{5}\)
Calculation of variance of u:

Question 8.
The mean of 5 observations is 4.8 and the variance is 6.56. If three of the five observations are 1, 3, and 8, find the other two observations.
Solution:
\(\bar{x}\) = 4.8, Var (X) = 6.56, n = 5, x₁ = 1, x₂ = 3, x₃ = 8 ……(given)
Let the remaining two observations be x₄ and x₅.


From (i), we get
x₅ = 5 or x₅ = 7
∴ The two numbers are 5 and 7.

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 8 Measures of Dispersion Ex 8.1 Questions and Answers.

11th Maths Part 1 Measures of Dispersion Exercise 8.1 Questions And Answers Maharashtra Board

Question 1.
Find a range of the following data:
19, 27, 15, 21, 33, 45, 7, 12, 20, 26
Solution:
Here, largest value (L) = 45, smallest value (S) = 7
∴ Range = L – S = 45 – 7 = 38

Question 2.
Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Solution:
Here, largest value (L) = 967, smallest value (S) = 250
∴ Range = L – S = 967 – 250 = 717

Question 3.
The following data gives a number of typing mistakes done by Radha during a week. Find the range of the data.

Solution:
Here, largest value (L) = 21, smallest value (S) = 10
∴ Range = L – S = 21 – 10 = 11

Question 4.
The following results were obtained by rolling a die 25 times. Find the range of the data.

Solution:
Here, largest value (L) = 6, smallest value (S) = 1
∴ Range = L – S = 6 – 1 = 5

Question 5.
Find range for the following data:

Solution:
Here, upper limit of the highest class (L) = 72
lower limit of the lowest class (S) = 62
∴ Range = L – S = 72 – 62 = 10

Class 11 Maharashtra State Board Maths Solution  

• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1
• Exercise 9.1 Class 11 Maths 1
• Exercise 9.2 Class 11 Maths 1
• Exercise 9.3 Class 11 Maths 1
• Exercise 9.4 Class 11 Maths 1
• Exercise 9.5 Class 11 Maths 1
• Miscellaneous Exercise 9 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Miscellaneous Exercise 7 Questions and Answers.

11th Maths Part 1 Conic Sections Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

(I) Select the correct option from the given alternatives.

Question 1.
The line y = mx + 1 is a tangent to the parabola y² = 4x, if m is ________
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(A) 1
Hint:
y² = 4x
Compare with y² = 4ax
∴ a = 1
Equation of tangent is y = mx + 1
Compare with y = mx + \(\frac{a}{m}\)
\(\frac{a}{m}\) = 1
∴ a = m = 1

Question 2.
The length of latus rectum of the parabola x² – 4x – 8y + 12 = 0 is ________
(A) 4
(B) 6
(C) 8
(D) 10
Answer:
(C) 8
Hint:
Given equation of parabola is
x² – 4x – 8y + 12 = 0
⇒ x² – 4x = 8y – 12
⇒ x² – 4x + 4 = 8y – 12 + 4
⇒ (x – 2)² = 8(y – 1)
Comparing this equation with (x – h)² = 4b(y – k), we get
4b = 8
∴ Length of latus rectum = 4b = 8

Question 3.
If the focus of the parabola is (0, -3), its directrix is y = 3, then its equation is ________
(A) x² = -12y
(B) x² = 12y
(C) y² = 12x
(D) y² = -12x
Answer:
(A) x² = -12y
Hint:

SP² = PM²
⇒ (x – 0)² + (y + 3)² = \(\left|\frac{y-3}{\sqrt{1}}\right|^{2}\)
⇒ x² + y² + 6y + 9 = y² – 6y + 9
⇒ x² = -12y

Question 4.
The co-ordinates of a point on the parabola y² = 8x whose focal distance is 4 are ________
(A) (\(\frac{1}{2}\), ±2)
(B) (1, ±2√2)
(C) (2, ±4)
(D) none of these
Answer:
(C) (2, ±4)

Question 5.
The end points of latus rectum of the parabola y² = 24x are ________
(A) (6, ±12)
(B) (12, ±6)
(C) (6, ±6)
(D) none of these
Answer:
(A) (6, ±12)

Question 6.
Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________
(A) y² = 8x
(B) y² = 32x
(C) y² = 16x
(D) x² = 32y
Answer:
(B) y² = 32x
Hint:
Since directrix is parallel to Y-axis,
The X-axis is the axis of the parabola.
Let the equation of parabola be y² = 4ax.
Equation of directrix is x + 8 = 0
∴ a = 8
∴ required equation of parabola is y² = 32x

Question 7.
The area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the endpoints of its latus rectum is ________
(A) 22 sq. units
(B) 20 sq. units
(C) 18 sq. units
(D) 14 sq. units
Answer:
(C) 18 sq. units
Hint:
x² = 12y
4b = 12
b = 3

Area of triangle = \(\frac{1}{2}\) × AB × OS
= \(\frac{1}{2}\) × 4a × a
= \(\frac{1}{2}\) × 12 × 3
= 18 sq. units

Question 8.
If P(\(\frac{\pi}{4}\)) is any point on the ellipse 9x² + 25y² = 225, S and S’ are its foci, then SP . S’P = ________
(A) 13
(B) 14
(C) 17
(D) 19
Answer:
(C) 17
Hint:
9x² + 25y² = 225
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Here, a = 5, b = 3
Eccentricity (e) = \(\frac{4}{5}\)
∴ \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{5}{\left(\frac{4}{5}\right)}=\frac{25}{4}\)
Coordinates of foci are S(4, 0) and S'(-4, 0)
P(θ) = (a cos θ, b sin θ)

Question 9.
The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________
(A) y² = 4x
(B) y² = 8x
(C) y² = -16x
(D) x² = 8y
Answer:
(B) y² = 8x
Hint:
The given points lie in the 1st and 4th quadrants.
∴ Equation of the parabola is y² = 4ax
End points of latus rectum are (a, 2a) and (a, -2a)
∴ a = 2
∴ required equation of parabola is y = 8x

Question 10.
If the parabola y² = 4ax passes through (3, 2), then the length of its latus rectum is ________
(A) \(\frac{2}{3}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1}{3}\)
(D) 4
Answer:
(B) \(\frac{4}{3}\)
Hint:
Length of latus rectum = 4a
The given parabola passes through (3, 2)
∴ (2)² = 4a(3)
∴ 4a = \(\frac{4}{3}\)

Question 11.
The eccentricity of rectangular hyperbola is
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{2^{\frac{1}{2}}}\)
(C) \(2^{\frac{1}{2}}\)
(D) \(\frac{1}{3^{\frac{1}{2}}}\)
Answer:
(C) \(2^{\frac{1}{2}}\)

Question 12.
The equation of the ellipse having one of the foci at (4, 0) and eccentricity \(\frac{1}{3}\) is
(A) 9x² + 16y² = 144
(B) 144x² + 9y² = 1296
(C) 128x² + 144y² = 18432
(D) 144x² + 128y² = 18432
Answer:
(C) 128x² + 144y² = 18432

Question 13.
The equation of the ellipse having eccentricity \(\frac{\sqrt{3}}{2}\) and passing through (-8, 3) is
(A) 4x² + y² = 4
(B) x² + 4y² = 100
(C) 4x² + y² = 100
(D) x² + 4y² = 4
Answer:
(B) x² + 4y² = 100

Question 14.
If the line 4x – 3y + k = 0 touches the ellipse 5x² + 9y² = 45, then the value of k is
(A) 21
(B) ±3√21
(C) 3
(D) 3(21)
Answer:
(B) ±3√21

Question 15.
The equation of the ellipse is 16x² + 25y² = 400. The equations of the tangents making an angle of 180° with the major axis are
(A) x = 4
(B) y = ±4
(C) x = -4
(D) x = ±5
Answer:
(B) y = ±4

Question 16.
The equation of the tangent to the ellipse 4x² + 9y² = 36 which is perpendicular to 3x + 4y = 17 is
(A) y = 4x + 6
(B) 3y + 4x = 6
(C) 3y = 4x + 6√5
(D) 3y = x + 25
Answer:
(C) 3y = 4x + 6√5

Question 17.
Eccentricity of the hyperbola 16x² – 3y² – 32x – 12y – 44 = 0 is
(A) \(\sqrt{\frac{17}{3}}\)
(B) \(\sqrt{\frac{19}{3}}\)
(C) \(\frac{\sqrt{19}}{3}\)
(D) \(\frac{\sqrt{17}}{3}\)
Answer:
(B) \(\sqrt{\frac{19}{3}}\)
Hint:
16x² – 3y² – 32x – 12y – 44 = 0
⇒ 16(x – 1)² – 3(y + 2)² = 48
⇒ \(\frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1\)
Here, a² = 3 and b² = 16
\(e=\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{3+16}}{\sqrt{3}}=\sqrt{\frac{19}{3}}\)

Question 18.
Centre of the ellipse 9x² + 5y² – 36x – 50y – 164 = 0 is at
(A) (2, 5)
(B) (1, -2)
(C) (-2, 1)
(D) (0, 0)
Answer:
(A) (2, 5)
Hint:
9x² + 5y² – 36x – 50y – 164 = 0
⇒ 9(x – 2)² + 5(y – 5)² = 325
⇒ \(\frac{(x-2)^{2}}{\frac{325}{9}}+\frac{(y-5)^{2}}{65}=1\)
⇒ centre of the ellipse = (2, 5)

Question 19.
If the line 2x – y = 4 touches the hyperbola 4x² – 3y² = 24, the point of contact is
(A) (1, 2)
(B) (2, 3)
(C) (3, 2)
(D) (-2, -3)
Answer:
(C) (3, 2)

Question 20.
The foci of hyperbola 4x² – 9y² – 36 = 0 are
(A) (±√13, 0)
(B) (±√11, 0)
(C) (±√12, 0)
(D) (0, ±√12)
Answer:
(A) (±√13, 0)

II. Answer the following.

Question 1.
For each of the following parabolas, find focus, equation of file directrix, length of the latus rectum and ends of the latus rectum.
(i) If 2y² = 17x
(ii) 5x² = 24y
Solution:
(i) Given equation of the parabola is 2y² = 17x
y² = \(\frac{17}{2}\)x
Comparing this equation with y² = 4ax, we get
4a = \(\frac{17}{2}\)
a = \(\frac{17}{8}\)
Co-ordinates of focus are S(a, 0), i.e., S(\(\frac{17}{8}\), 0)
Equation of the directrix is x + a = 0
x + \(\frac{17}{8}\) = 0
8x + 17 = 0
Length of latus rectum = 4a = 4(\(\frac{17}{8}\)) = \(\frac{17}{2}\)
Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a)
i.e., \(\left(\frac{17}{8}, \frac{17}{4}\right)\) and \(\left(\frac{17}{8},-\frac{17}{4}\right)\)

(ii) Given equation of the parabola is 5x² = 24y
x² = \(\frac{24 y}{5}\)
Comparing this equation with x² = 4by, we get
4b = \(\frac{24}{5}\)
b = \(\frac{6}{5}\)
Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{6}{5}\))
Equation of the directrix is y + b = 0
y + \(\frac{6}{5}\) = 0
5y + 6 = 0
Length of latus rectum = 4b = 4(\(\frac{6}{5}\)) = \(\frac{24}{5}\)
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b), i.e., \(\left(\frac{12}{5}, \frac{6}{5}\right)\) and \(\left(\frac{-12}{5}, \frac{6}{5}\right)\)

Question 2.
Find the cartesian co-ordinates of the points on the parabola y² = 12x whose parameters are
(i) 2
(ii) -3
Solution:
Given equation of the parabola is y² = 12x
Comparing this equation with y² = 4ax, we get
4a = 12
∴ a = 3
If t is the parameter of the point P on the parabola, then
P(t) = (at², 2at)
i.e., x = at² and y = 2at …..(i)
(i) Given, t = 2
Substituting a = 3 and t = 2 in (i), we get
x = 3(2)² and y = 2(3)(2)
x = 12 and y = 12
∴ The cartesian co-ordinates of the point on the parabola are (12, 12).

(ii) Given, t = -3
Substitùting a = 3 and t = -3 in (i), we get
x = 3(-3)² and y = 2(3)(-3)
∴ x = 27 and y = -18
∴ The cartesian co-ordinates of the point on the parabola are (27, -18).

Question 3.
Find the co-ordinates of a point of the parabola y² = 8x having focal distance 10.
Solution:
Given equation of the parabola is y² = 8x
Comparing this equation with y² = 4ax, we get
4a = 8
∴ a = 2
Focal distance of a point = x + a
Given, focal distance = 10
x + 2 = 10
∴ x = 8
Substituting x = 8 in y² = 8x, we get
y² = 8(8)
∴ y = ±8
∴ The co-ordinates of the points on the parabola are (8, 8) and (8, -8).

Question 4.
Find the equation of the tangent to the parabola y² = 9x at the point (4, -6) on it.
Solution:
Given equation of the parabola is y² = 9x
Comparing this equation with y² = 4ax, we get
4a = 9
∴ a = \(\frac{9}{4}\)
Equation of the tangent y² = 4ax at (x₁, y₁) is yy₁ = 2a(x + x₁)
The equation of the tangent at (4, -6) is
y(-6) = 2(\(\frac{9}{4}\))(x + 4)
⇒ -6y = \(\frac{9}{2}\) (x + 4)
⇒ -12y = 9x + 36
⇒ 9x + 12y + 36 = 0
⇒ 3x + 4y + 12 = 0

Question 5.
Find the equation of the tangent to the parabola y² = 8x at t = 1 on it.
Solution:
Given equation of the parabola is y² = 8x
Comparing this equation with y² = 4ax, we get
4a = 8
a = 2
t = 1
Equation of tangent with parameter t is yt = x + at²
∴ The equation of tangent with t = 1 is
y(1) = x + 2(1)²
y = x + 2
∴ x – y + 2 = 0

Question 6.
Find the equations of the tangents to the parabola y² = 9x through the point (4, 10).
Solution:
Given equation of the parabola is y² = 9x
Comparing this equation with y² = 4ax, we get
4a = 9
∴ a = \(\frac{9}{4}\)
Equation of tangent to the parabola y² = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
y = mx + \(\frac{9}{4 m}\)
But, (4, 10) lies on the tangent.
10 = 4m + \(\frac{9}{4 m}\)
⇒ 40m = 16m²+ 9
⇒ 16m² – 40m + 9 = 0
⇒ 16m² – 36m – 4m + 9 = 0
⇒ 4m(4m – 9) – 1(4m – 9) = 0
⇒ (4m – 9) (4m – 1) = 0
⇒ 4m – 9 = 0 or 4m – 1 = 0
⇒ m = \(\frac{9}{4}\) or m = \(\frac{1}{4}\)
These are the slopes of the required tangents.
By slope point form, y – y₁ = m(x – x₁),
the equations of the tangents are
y – 10 = \(\frac{9}{4}\)(x – 4) or y – 10 = \(\frac{1}{4}\)(x – 4)
⇒ 4y – 40 = 9x – 36 or 4y – 40 = x – 4
⇒ 9x – 4y + 4 = 0 or x – 4y + 36 = 0

Question 7.
Show that the two tangents drawn to the parabola y² = 24x from the point (-6, 9) are at the right angle.
Solution:
Given the equation of the parabola is y² = 24x.
Comparing this equation with y² = 4ax, we get
4a = 24
⇒ a = 6
Equation of tangent to the parabola y² = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
⇒ y = mx + \(\frac{6}{m}\)
But, (-6, 9) lies on the tangent
9 = -6m + \(\frac{6}{m}\)
⇒ 9m = -6m² + 6
⇒ 6m² + 9m – 6 = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁m₂ = -1
Tangents drawn to the parabola y² = 24x from the point (-6, 9) are at a right angle.

Alternate method:
Comparing the given equation with y² = 4ax, we get
4a = 24
⇒ a = 6
Equation of the directrix is x = -6.
The given point lies on the directrix.
Since tangents are drawn from a point on the directrix are perpendicular,
Tangents drawn to the parabola y² = 24x from the point (-6, 9) are at the right angle.

Question 8.
Find the equation of the tangent to the parabola y² = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.
Solution:
Given the equation of the parabola is y² = 8x.
Comparing this equation with y² = 4ax, we get
4a = 8
a = 2
Slope of the line 2x + 2y + 5 = 0 is -1
Since the tangent is parallel to the given line,
slope of the tangent line is m = -1
Equation of tangent to the parabola y² = 4ax having slope m is y = mx + \(\frac{a}{m}\)
Equation of the tangent is
y = -x + \(\frac{2}{-1}\)
x + y + 2 = 0
Point of contact = \(\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)\)
= \(\left(\frac{2}{(-1)^{2}}, \frac{2(2)}{-1}\right)\)
= (2, -4)

Question 9.
A line touches the circle x² + y² = 2 and the parabola y² = 8x. Show that its equation is y = ±(x + 2).
Solution:
Given equation of the parabola is y² = 8x
Comparing this equation with y² = 4ax, we get
4a = 8
a = 2
Equation of tangent to given parabola with slope m is
y = mx + \(\frac{2}{m}\)
m²x – my + 2 = 0 ….(i)
Equation of the circle is x² + y² = 2
Its centre = (0, 0) and Radius = √2
Line (i) touches the circle.
Length of perpendicular from the centre to the line (i) = radius
⇒ \(\left|\frac{m^{2}(0)-m(0)+2}{\sqrt{m^{4}+m^{2}}}\right|\) = √2
⇒ \(\frac{4}{m^{4}+m^{2}}\) = 2
⇒ m⁴ + m² – 2 – 0
⇒ (m² + 2)(m² – 1) = 0
Since m² ≠ -2,
m² – 1 = 0
⇒ m = ±1
When m = 1, equation of the tangent is
y = (1)x + \(\frac{2}{(1)}\)
y = (x + 2) …..(i)
When m = -1, equation of the tangent is
y = (-1)x + \(\frac{2}{(-1)}\)
y = -x – 2
y = -(x + 2) …..(ii)
From (i) and (ii),
equation of the common tangents to the given parabola is y = ±(x + 2)

Question 10.
Two tangents to the parabola y² = 8x meet the tangents at the vertex in P and Q. If PQ = 4, prove that the locus of the point of intersection of the two tangents is y² = 8(x + 2).
Solution:
Given parabola is y² = 8x
Comparing with y² = 4ax, we get,
4a = 8
⇒ a = 2
Let M(t₁) and N(t₂) be any two points on the parabola.
The equations of tangents at M and N are
yt₁ = x + \(2 \mathrm{t}_{1}^{2}\) …..(1)
yt₂ = x + \(2 \mathrm{t}_{2}^{2}\) …(2) ….[∵ a = 2]
Let tangent at M meet the tangent at the vertex in P.
But tangent at the vertex is Y-axis whose equation is x = 0.
⇒ to find P, put x = 0 in (1)
⇒ yt₁ = \(2 \mathrm{t}_{1}^{2}\)
⇒ y = 2t₁ …..(t₁ ≠ 0 otherwise tangent at M will be x = 0)
⇒ P = (0, 2t₁)
Similarly, Q = (0, 2t₂)
It is given that PQ = 4
∴ |2t₁ – 2t₂| = 4
∴ |t₁ – t₂| = 2 …..(3)
Let R = (x₁, y₁) be any point on the required locus.
Then R is the point of intersection of tangents at M and N.
To find R, we solve (1) and (2).
Subtracting (2) from (1), we get
y(t₁ – t₂) = \(2 \mathrm{t}_{1}^{2}-2 \mathrm{t}_{2}^{2}\)
y(t₁ – t₂) = 2(t₁ – t₂)(t₁ + t₂)
∴ y = 2(t₁ + t₂) …..[∵ M, N are distinct ∴ t₁ ≠ t₂]
i.e., y₁ = 2(t₁ + t₂) …..(4)
∴ from (1), we get
2t₁(t₁ + t₂) = x + \(2 \mathrm{t}_{1}^{2}\)
∴ 2t₁t₂ = x i.e. x₁ = 2t₁t₂ …..(5)
To find the equation of locus of R(x₁, y₁),
we eliminate t₁ and t₂ from the equations (3), (4) and (5).
We know that,
(t₁ + t₂)² = (t₁ + t₂)² + 4t₁t₂
⇒ \(\left(\frac{y_{1}}{2}\right)^{2}=4+4\left(\frac{x_{1}}{2}\right)\) …[By (3), (4) and (5)]
⇒ \(y_{1}^{2}\) = 16 + 8x₁ = 8(x₁ + 2)
Replacing x₁ by x and y₁ by y,
the equation of required locus is y² = 8(x + 2).

Question 11.
The slopes of the tangents drawn from P to the parabola y² = 4ax are m₁ and m₂, showing that
(i) m₁ – m₂ = k
(ii) \(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.
Solution:
Let P(x₁, y₁) be any point on the parabola y² = 4ax.
Equation of tangent to the parabola y² = 4ax having slope m is y = mx + \(\frac{\mathrm{a}}{\mathrm{m}}\)
This tangent passes through P(x₁, y₁).
y₁ = mx₁ + \(\frac{\mathrm{a}}{\mathrm{m}}\)
my₁ = m²x₁ + a
m²x₁ – my₁ + a = 0
This is a quadratic equation in ‘m’.
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents drawn from P.
∴ m₁ + m₂ = \(\frac{y_{1}}{x_{1}}\), m₁m₂ = \(\frac{a}{x_{1}}\)

Since (x₁, y₁) and a are constants, m₁ – m₂ is a constant.
∴ m₁ – m₂ = k, where k is constant.

(ii) Since (x₁, y₁) and a are constants, m₁m₂ is a constant.
\(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.

Question 12.
The tangent at point P on the parabola y² = 4ax meets the Y-axis in Q. If S is the focus, show that SP subtends a right angle at Q.
Solution:
Let P(\(a t_{1}^{2}\), 2at₁) be a point on the parabola and
S(a, 0) be the focus of parabola y² = 4ax

Since the tangent passing through point P meet Y-axis at point Q,
equation of tangent at P(\(a t_{1}^{2}\), 2at₁) is
yt₁ = x + \(a t_{1}^{2}\) …..(i)
∴ Point Q lie on tangent
∴ put x = 0 in equation (i)
yt₁ = \(a t_{1}^{2}\)
y = at₁
∴ Co-ordinate of point Q(0, at₁)
S = (a, 0), P(\(a t_{1}^{2}\), 2at₁), Q(0, at₁)

∴ SP subtends a right angle at Q.

Question 13.
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) Distance between foci
(vi) distance between directrices of the curve
(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
(b) 16x² + 25y² = 400
(c) \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)
(d) x² – y² = 16
Solution:
(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 9
∴ a = 5 and b = 3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
∴ Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
∴ e = \(\frac{\sqrt{25-9}}{5}\) = \(\frac{4}{5}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)
i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0),
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
i.e., x = ±\(\frac{5}{\frac{4}{5}}\)
i.e., x = ±\(\frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae = 2 (5) (\(\frac{4}{5}\)) = 8

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(5)}{\frac{4}{5}}\) = \(\frac{25}{2}\)

(b) Given equation of the ellipse is 16x² + 25y² = 400
\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 16
∴ a = 5 and b = 4
Since a > b,
X-axis is the major axis and Y-axis is the minor axis
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(4) = 8
Lengths of the principal axes are 10 and 8.

(ii) b² = a² (1 – e²)
16 = 25(1 – e²)
\(\frac{16}{25}\) = 1 – e²
e² = 1 – \(\frac{16}{25}\)
e² = \(\frac{9}{25}\)
e = \(\frac{3}{5}\) ……[∵ 0 < e < 1]
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(\(\frac{3}{5}\)), 0) and S'(-5(\(\frac{3}{5}\)), 0),
i.e., S(3, 0) and S'(-3, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
i.e., x = ±\(\frac{5}{\left(\frac{3}{5}\right)}\)
i.e., x = ±\(\frac{25}{3}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{5}=\frac{32}{5}\)

(v) Distance between foci = 2ae = 2(5)(\(\frac{3}{5}\)) = 6

(vi) Distance between directrices = \(\frac{2 a}{e}=\frac{2(5)}{\left(\frac{3}{5}\right)}=\frac{50}{3}\)

(c) Given equation of the hyperbola \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
a² = 144 and b² = 25
∵ a = 12 and b = 5
(i) Length of transverse axis = 2a = 2(12) = 24
Length of conjugate axis = 2b = 2(5) = 10
lengths of the principal axes are 24 and 10.

(ii) b² = a²(e² – 1)
25 = 144 (e² – 1)
\(\frac{25}{144}\) = e² – 1
e² = 1 + \(\frac{25}{144}\)
e² = \(\frac{169}{144}\)
e = \(\frac{13}{12}\) …….[∵ e > 1]
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0)
i.e., S(12(\(\frac{13}{12}\)), 0) and S'(-12(\(\frac{13}{12}\)), 0)
i.e., S(13, 0) and S'(-13, 0)

(iii) Equations of the directrices are x = \(\pm \frac{a}{e}\)
i.e., x = \(\pm \frac{12}{\left(\frac{13}{12}\right)}\)
i.e., x = \(\pm \frac{144}{13}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(25)}{12}=\frac{25}{6}\)

(v) Distance between foci = 2ae = 2(12)(\(\frac{13}{12}\)) = 26

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(12)}{\left(\frac{13}{12}\right)}\) = \(\frac{288}{13}\)

(d) Given equation of the hyperbola is x² – y² = 16
∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 16
∴ a = 4 and b = 4
(i) Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8

(ii) We know that

Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(4√2, 0) and S'(-4√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)
∴x = ± \(\frac{4}{\sqrt{2}}\)
∴ x = ±2√2

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(16)}{4}\) = 8

(v) Distance between foci = 2ae = 2(4)(√2) = 8√2

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(4)}{\sqrt{2}}\) = 4√2.

Question 14.
Find the equation of the ellipse in standard form if
(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.
(ii) the length of the major axis is 10 and the distance between foci is 8.
(iii) passing through the points (-3, 1) and (2, -2).
Solution:
(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{3}{8}\)
Distance between foci = 2ae
Given, distance between foci = 6
∴ 2ae = 6
∴ 2a(\(\frac{3}{8}\)) = 6
∴ \(\frac{3a}{4}\) = 6
∴ a = 8
∴ a² = 64
Now, b² = a² (1 – e²)
= \(64\left[1-\left(\frac{3}{8}\right)^{2}\right]\)
= \(4\left(1-\frac{9}{64}\right)\)
= 64(\(\frac{55}{64}\))
= 55
∴ The required equation of the ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\)

(ii) Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) ……(1)
Then length of major axis = 2a = 10
∴ a = 5
Also, distance between foci= 2ae = 8
∴ 2 × 5 × e = 8
∴ e = \(\frac{4}{5}\)
∴ b² = a²(1 – e²)
= 25(1 – \(\frac{6}{25}\))
= 9
∴ from (1), the equation of the required ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
∴ Substituting x = -3 and y = 1 in equation of ellipse, we get
\(\frac{(-3)^{2}}{a^{2}}+\frac{1^{2}}{b^{2}}=1\)
∴ \(\frac{9}{a^{2}}+\frac{1}{b^{2}}=1\) …..(i)
Substituting x = 2 and y = -2 in equation of ellipse, we get
\(\frac{2^{2}}{a^{2}}+\frac{(-2)^{2}}{b^{2}}=1\)
∴ \(\frac{4}{a^{2}}+\frac{4}{b^{2}}=1\) ……(ii)
Let \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B
∴ Equations (i) and (ii) become
9A + B = 1 ..…(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
∴ A = \(\frac{3}{32}\)
Substituting A = \(\frac{3}{32}\) in (iv), we get
4(\(\frac{3}{32}\)) + 4B = 1
∴ \(\frac{3}{8}\) + 4B = 1
∴ 4B = 1 – \(\frac{3}{8}\)
∴ 4B = \(\frac{5}{8}\)
∴ B = \(\frac{5}{32}\)
Since \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B
\(\frac{1}{a^{2}}=\frac{3}{32}\) and \(\frac{1}{b^{2}}=\frac{5}{32}\)
∴ a² = \(\frac{32}{3}\) and b² = \(\frac{32}{5}\)
∴ The required equation of ellipse is
\(\frac{x^{2}}{\left(\frac{32}{3}\right)}+\frac{y^{2}}{\left(\frac{32}{5}\right)}\)
i.e., 3x² + 5y² = 32.

Question 15.
Find the eccentricity of an ellipse if the distance between its directrices is three times the distance between its foci.
Solution:
Let the equation of the ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
It is given that,
distance between directrices is three times the distance between the foci.
∴ \(\frac{2a}{e}\) = 3(2ae)
∴ 1 = 3e²
∴ e² = \(\frac{1}{3}\)
∴ e = \(\frac{1}{\sqrt{3}}\) …..[∵ 0 < e < 1]

Question 16.
For the hyperbola \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\), prove that SA . S’A = 25, where S and S’ are the foci and A is the vertex.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 100 and b² = 25
∴ a = 10 and b = 5
∴ Co-ordinates of vertex is A(a, 0), i.e., A(10, 0)
Eccentricity, e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}\)
= \(\frac{\sqrt{100+25}}{10}\)
= \(\frac{\sqrt{125}}{10}\)
= \(\frac{5 \sqrt{5}}{10}\)
= \(\frac{\sqrt{5}}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)
i.e., S(10(\(\frac{\sqrt{5}}{2}\)), 0) and S'(-10(\(\frac{\sqrt{5}}{2}\)), 0)
i.e., S(5√5, 0) and S'(-5√5, 0)
Since S, A and S’ lie on the X-axis,
SA = |5√5 – 10| and S’A = |-5√5 – 10|
= |-(5√5 + 10)|
= |5√5 + 10|
∴ SA . S’A = |5√5 – 10| |5√5 + 10|
= |(5√5)² – (10)²|
= |125 – 100|
= |25|
SA . S’A = 25

Question 17.
Find the equation of the tangent to the ellipse \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 5 and b² = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, -2) lies on both the tangents,
-2 = 2m ± \(\sqrt{5 m^{2}+4}\)
∴ -2 – 2m = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
4m² + 8m + 4 = 5m² + 4
∴ m² – 8m = 0
∴ m(m – 8) = 0
∴ m = 0 or m = 8
These are the slopes of the required tangents.
∴ By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y + 2 = 0(x – 2) and y + 2 = 8(x – 2)
∴ y + 2 = 0 and y + 2 = 8x – 16
∴ y + 2 = 0 and 8x – y – 18 = 0.

Question 18.
Find the equation of the tangent to the ellipse x² + 4y² = 100 at (8, 3).
Solution:
Given equation of ellipse is x² + 4y² = 100
∴ \(\frac{x^{2}}{100}+\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 100 and b² = 25
Equation of tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at (x₁, y₁) is \(\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1\)
Equation of tangent at (8, 3) is
\(\frac{8 x}{100}+\frac{3 y}{25}=1\)
\(\frac{2 x}{25}+\frac{3 y}{25}=1\)
2x + 3y = 25

Question 19.
Show that the line 8y + x = 17 touches the ellipse x² + 4y² = 17. Find the point of contact.
Solution:

Question 20.
Tangents are drawn through a point P to the ellipse 4x² + 5y² = 20 having inclinations θ₁ and θ₂ such that tan θ₁ + tan θ₂ = 2. Find the equation of the locus of P.
Solution:
Given equation of the ellipse is 4x² + 5y² = 20.
∴ \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 5 and b² = 4
Since inclinations of tangents are θ₁ and θ₂,
m₁ = tan θ₁ and m₂ = tan θ₂
Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)
∴ y = mx ± \(\sqrt{5 m^{2}+4}\)
∴ y – mx = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
y² – 2mxy + m²x² = 5m² + 4
∴ (x² – 5)m² – 2xym + (y² – 4) = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
∴ m₁ + m₂ = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)
Given, tan θ₁ + tan θ₂ = 2
∴ m₁ + m₂ = 2
∴ \(\frac{2 x y}{x^{2}-5}\)
∴ xy = x² – 5
∴ x² – xy – 5 = 0, which is the required equation of the locus of P.

Question 21.
Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
∴ a² = 25, b² = 16
∴ a = 5, b = 4
We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
∴ e = \(\frac{\sqrt{25-16}}{5}\) = \(\frac{3}{5}\)
ae = 5(\(\frac{3}{5}\)) = 3
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3, 0) and S'(-3, 0)
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)
Equation of one of the tangents to the ellipse is
y = mx + \(\sqrt{25 \mathrm{~m}^{2}+16}\)
∴ mx – y + \(\sqrt{25 \mathrm{~m}^{2}+16}\) = 0 …..(i)
p₁ = length of perpendicular segment from S(3, 0) to the tangent (i)

p₂ = length of perpendicular segment from S'(-3, 0) to the tangent (i)

Question 22.
Find the equation of the hyperbola in the standard form if
(i) Length of conjugate axis is 5 and distance between foci is 13.
(ii) eccentricity is \(\frac{3}{2}\) and distance between foci is 12.
(iii) length of the conjugate axis is 3 and the distance between the foci is 5.
Solution:
(i) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 5
2b = 5
b = \(\frac{5}{2}\)
b² = \(\frac{25}{4}\)
Distance between foci = 2ae
Given, distance between foci = 13
2ae = 13
ae = \(\frac{13}{2}\)
a²e² = \(\frac{169}{4}\)
Now, b² = a²(e² – 1)
b² = a²e² – a²
\(\frac{25}{4}\) = \(\frac{169}{4}\) – a²
a² = \(\frac{169}{4}-\frac{25}{4}\) = 36
∴ The required equation of hyperbola is \(\frac{x^{2}}{36}-\frac{y^{2}}{\frac{25}{4}}=1\)
i.e., \(\frac{x^{2}}{36}-\frac{4 y^{2}}{25}=1\)

(ii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Given, eccentricity (e) = \(\frac{3}{2}\)
Distance between foci = 2ae
Given, distance between foci = 12
∴ 2ae = 12
∴ 2a(\(\frac{3}{2}\)) = 12
∴ 3a = 12
∴ a = 4
∴ a² = 16
Now, b² = a²(e² – 1)
∴ b² = \(\left[\left(\frac{3}{2}\right)^{2}-1\right]\)
∴ b² = 16(\(\frac{9}{4}\) – 1)
∴ b² = 16(\(\frac{5}{4}\))
∴ b² = 20
∴ The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 3
∴ 2b = 3
∴ b = \(\frac{3}{2}\)
∴ b² = \(\frac{9}{4}\)
Distance between foci = 2ae
Given, distance between foci = 5
∴ 2ae = 5
∴ ae = \(\frac{5}{2}\)
∴ a²e² = \(\frac{25}{4}\)
Now, b² = a²(e² – 1)
∴ b² = a²e² – a²
∴ \(\frac{9}{4}\) = \(\frac{25}{4}\) – a²
∴ a² = \(\frac{25}{4}-\frac{9}{4}\)
∴ a² = 4
∴ The required equation of hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{\left(\frac{9}{4}\right)}=1\)
i.e., \(\frac{x^{2}}{4}-\frac{4 y^{2}}{9}=1\)

Question 23.
Find the equation of the tangent to the hyperbola,
(i) 7x² – 3y² = 51 at (-3, -2)
(ii) x = 3 sec θ, y = 5 tan θ at θ = π/3
(iii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\) at P(30°).
Solution:
(i) Given equation of the hyperbola is 7x² – 3y² = 51

(ii) Given, equation of the hyperbola is
x = 3 sec θ, y = 5 tan θ
Since sec² θ – tan² θ = 1,
\(\frac{x^{2}}{9}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 9 and b² = 25
a = 3 and b = 5
Equation of tangent at P(θ) is
\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)
∴ Equation of tangent at P(π/3) is
\(\frac{x \sec \left(\frac{\pi}{3}\right)}{3}-\frac{y \tan \left(\frac{\pi}{3}\right)}{5}=1\)
\(\frac{2 x}{3}-\frac{\sqrt{3} y}{5}=1\)
10x – 3√3 y = 15

(iii) Given equation of hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 16
a = 5 and b = 4
Equation of tangent at P(θ) is
\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)
The equation of tangent at P(30°) is
\(\frac{x \sec 30^{\circ}}{5}-\frac{y \tan 30^{\circ}}{4}=1\)
\(\frac{2 x}{5 \sqrt{3}}-\frac{y}{4 \sqrt{3}}=1\)
8x – 5y = 20√3

Question 24.
Show that the line 2x – y = 4 touches the hyperbola 4x² – 3y² = 24. Find the point of contact.
Solution:
Given equation of die hyperbola is 4x² – 3y² = 24.
∴ \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 6 and b² = 8
Given equation of line is 2x – y = 4
∴ y = 2x – 4
Comparing this equation with y = mx + c, we get
m = 2 and c = -4
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have
c² = a²m² – b²
c² = (-4)² = 16
a²m² – b² = 6(2)² – 8 = 24 – 8 = 16
∴ The given line is a tangent to the given hyperbola and point of contact
= \(\left(-\frac{\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}},-\frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{-6(2)}{-4}, \frac{-8}{-4}\right)\)
= (3, 2)

Question 25.
Find the equations of the tangents to the hyperbola 3x² – y² = 48 which are perpendicular to the line x + 2y – 7 = 0.
Solution:
Given the equation of the hyperbola is 3x² – y² = 48.
∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{48}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 48
Slope of the line x + 2y – 7 = 0 is \(-\frac{1}{2}\)
Since the given line is perpendicular to the tangents,
slope of the required tangent (m) = 2
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Equations of tangents to the ellipse having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = 2x ± \(\sqrt{16(2)^{2}-48}\)
y = 2x ± √16
∴ y = 2x ± 4

Question 26.
Two tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) make angles θ₁, θ₂, with the transverse axis. Find the locus of their point of intersection if tan θ₁ + tan θ₂ = k.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Let θ₁ and θ₂ be the inclinations.
m₁ = tan θ₁, m₂ = tan θ₂
Let P(x₁, y₁) be a point on the hyperbola
Equation of a tangent with slope ‘m’ to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P(x₁, y₁).
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
\(\left(x_{1}{ }^{2}-\mathrm{a}^{2}\right) \mathrm{m}^{2}-2 x_{1} y_{1} \mathrm{~m}+\left(y_{1}{ }^{2}+\mathrm{b}^{2}\right)=0\) ……(i)
This is a quadratic equation in ‘m’.
It has two roots say m₁ and m₂, which are the slopes of two tangents drawn from P.
∴ m₁ + m₂ = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\)
Since tan θ₁ + tan θ₂ = k,
\(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=k\)
∴ P(x₁, y₁) moves on the curve whose equation is k(x² – a²) = 2xy.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.3 Questions and Answers.

11th Maths Part 1 Conic Sections Exercise 7.3 Questions And Answers Maharashtra Board

Question 1.
Find the length of the transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices, and the length of the latus rectum of the hyperbolae.
(i) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
(ii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1\)
(iii) 16x² – 9y² = 144
(iv) 21x² – 4y² = 84
(v) 3x² – y² = 4
(vi) x² – y² = 16
(vii) \(\frac{y^{2}}{25}-\frac{x^{2}}{9}=1\)
(viii) \(\frac{y^{2}}{25}-\frac{x^{2}}{144}=1\)
(ix) \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
(x) x = 2 sec θ, y = 2√3 tan θ
Solution:
(i) Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 16
⇒ a = 5 and b = 4
Length of transverse axis = 2a = 2(5) = 10
Length of conjugate axis = 2b = 2(4) = 8
We know that

(ii) Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1\)
\(\frac{y^{2}}{16}-\frac{x^{2}}{25}=1\)
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b² = 16 and a² = 25
⇒ b = 4 and a = 5
Length of transverse axis = 2b = 2(4) = 8
Length of conjugate axis = 2a = 2(5) = 10
Co-ordinates of vertices are B(0, b) and B’ (0, -b)
i.e., B(0, 4) and B'(0, -4)
We know that

(iii) Given equation of the hyperbola is 16x² – 9y² = 144.
\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 9 and b² = 16
⇒ a = 3 and b = 4
Length of transverse axis = 2a = 2(3) = 6
Length of conjugate axis = 2b = 2(4) = 8
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{9+16}}{3}=\frac{\sqrt{25}}{3}=\frac{5}{3}\)
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3(\(\frac{5}{3}\)), 0) and S'(-3(\(\frac{5}{3}\)), 0)
i.e., S(5, 0) and S'(-5, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\)
= \(\pm \frac{3}{\left(\frac{5}{3}\right)}\)
= \(\pm \frac{9}{5}\)
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{3}=\frac{32}{3}\)

(iv) Given equation of the hyperbola is 21x² – 4y² = 84.
\(\frac{x^{2}}{4}-\frac{y^{2}}{21}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 4 and b² = 21
⇒ a = 2 and b = √21
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2√21
We know that

(v) Given equation of the hyperbola is 3x² – y² = 4.
\(\frac{x^{2}}{\left(\frac{4}{3}\right)}-\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = \(\frac{4}{3}\) and b² = 4

(vi) Given equation of the hyperbola is x² – y² = 16.
\(\frac{x^{2}}{16}-\frac{y^{2}}{16}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 16
⇒ a = 4 and b = 4
Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{16+16}}{4}=\frac{\sqrt{32}}{4}=\frac{4 \sqrt{2}}{4}=\sqrt{2}\)
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S (4√2, 0) and S’ (-4√2, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\)
⇒ x = \(\pm \frac{4}{\sqrt{2}}\)
⇒ x = ± 2√2
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{4}\) = 8

(vii) Given equation of the hyperbola is \(\frac{y^{2}}{25}-\frac{x^{2}}{9}=1\).
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b² = 25 and a² = 9
⇒ b = 5 and a = 3
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(3) = 6
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that

(viii) Given equation of the hyperbola is \(\frac{y^{2}}{25}-\frac{x^{2}}{144}=1\).
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b² = 25 and a² = 144
⇒ b = 5 and a = 12
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(12) = 24
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that

(ix) Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 100 and b² = 25
⇒ a = 10 and b = 5
Length of transverse axis = 2a = 2(10) = 20
Length of conjugate axis = 2b = 2(5) = 10
We know that

(x) Given equation of the hyperbola is x = 2 sec θ, y = 2√3 tan θ.
Since sec² θ – tan² θ = 1,
\(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2 \sqrt{3}}\right)^{2}=1\)
\(\frac{x^{2}}{4}-\frac{y^{2}}{12}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 4 and b² = 12
⇒ a = 2 and b = 2√3
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2(2√3) = 4√3
We know that
e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}\) = \(\frac{\sqrt{4+12}}{2}\) = 2
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(2), 0) and S'(-2(2), 0),
i.e., S(4, 0) and S'(-4, 0)
Equations of the directrices are x = ±\(\frac{a}{e}\).
⇒ x = ±\(\frac{2}{2}\)
⇒ x = ±1
Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(12)}{2}\) = 12

Question 2.
Find the equation of the hyperbola with centre at the origin, length of the conjugate axis as 10, and one of the foci as (-7, 0).
Solution:
Given, one of the foci of the hyperbola is (-7, 0).
Since this focus lies on the X-axis, it is a standard hyperbola.
Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 10
⇒ 2b = 10
⇒ b = 5
⇒ b² = 25
Co-ordinates of focus are (-ae, 0)
ae = 7
⇒ a²e² = 49
Now, b² = a²(e² – 1)
⇒ 25 = 49 – a²
⇒ a² = 49 – 25 = 24
The required equation of hyperbola is \(\frac{x^{2}}{24}-\frac{y^{2}}{25}=1\)

Question 3.
Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x² – 3y² = 3
Solution:
Given, equation of hyperbola is x² – 3y² = 3.
\(\frac{x^{2}}{3}-\frac{y^{2}}{1}=1\)
Equation of the hyperbola conjugate to the above hyperbola is \(\frac{y^{2}}{1}-\frac{x^{2}}{3}=1\)
Comparing this equation with \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\), we get
b² = 1 and a² = 3
Now, a² = b²(e² – 1)
⇒ 3 = 1(e² – 1)
⇒ 3 = e – 1
⇒ e² = 4
⇒ e = 2 …..[∵ e > 1]

Question 4.
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that \(\frac{1}{e^{2}}+\frac{1}{\left(e^{\prime}\right)^{2}}=1\).
Solution:
Let e be the eccentricity of a hyperbola

Question 5.
Find the equation of the hyperbola referred to its principal axes:
(i) whose distance between foci is 10 and eccentricity is \(\frac{5}{2}\)
(ii) whose distance between foci is 10 and length of the conjugate axis is 6.
(iii) whose distance between directrices is \(\frac{8}{3}\) and eccentricity is \(\frac{3}{2}\).
(iv) whose length of conjugate axis = 12 and passing through (1, -2).
(v) which passes through the points (6, 9) and (3, 0).
(vi) whose vertices are (±7, 0) and endpoints of the conjugate axis are (0, ±3).
(vii) whose foci are at (±2, 0) and eccentricity is \(\frac{3}{2}\).
(viii) whose lengths of transverse and conjugate axes are 6 and 9 respectively.
(ix) whose length of transverse axis is 8 and distance between foci is 10.
Solution:
(i) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Given, eccentricity (e) = \(\frac{5}{2}\)
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a(\(\frac{5}{2}\)) = 5
⇒ a = 2
⇒ a² = 4

(ii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of conjugate axis = 2b
Given, length of conjugate axis = 6
⇒ 2b = 6
⇒ b = 3
⇒ b² = 9
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a²e² = 25
Now, b² = a² (e² – 1)
⇒ b² = a² e² – a²
⇒ 9 = 25 – a²
⇒ a² = 25 – 9
⇒ a² = 16
The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\)
Given, eccentricity (e) = \(\frac{3}{2}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{8}{3}\)

(iv) Let the required equation of hyperbola be
\(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\) ……(i)
Length of conjugate axis = 2b
Given, length of conjugate axis = 12
⇒ 2b = 12
⇒ b = 6 …..(ii)
⇒ b² = 36
The hyperbola passes through (1, -2)
Substituting x = 1 and y = -2 in (i), we get

(v) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) ……(i)
The hyperbola passes through the points (6, 9) and (3, 0).

(vi) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Co-ordinates of vertices are (±a, 0).
Given that, co-ordinates of vertices are (±7, 0)
∴ a = 7
Endpoints of the conjugate axis are (0, b) and (0, -b).
Given, the endpoints of the conjugate axis are (0, ±3).
∴ b = 3
The required equation of hyperbola is \(\frac{x^{2}}{7^{2}}-\frac{y^{2}}{3^{2}}=1\)
i.e., \(\frac{x^{2}}{49}-\frac{y^{2}}{9}=1\)

(vii) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) ……(i)
Given, eccentricity (e) = \(\frac{3}{2}\)
Co-ordinates of foci are (±ae, 0).
Given co-ordinates of foci are (±2, 0)
ae = 2
⇒ a(\(\frac{3}{2}\)) = 2
⇒ a = \(\frac{4}{3}\)
⇒ a² = \(\frac{16}{9}\)

(viii) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of transverse axis = 2a
Given, length of transverse axis = 6
⇒ 2a = 6
⇒ a = 3
⇒ a² = 9
Length of conjugate axis = 2b
Given, length of conjugate axis = 9
⇒ 2b = 9
⇒ b = \(\frac{9}{2}\)
⇒ b² = \(\frac{81}{4}\)
The required equation of hyperbola is
\(\frac{x^{2}}{9}-\frac{y^{2}}{\left(\frac{81}{4}\right)}=1\)
i.e., \(\frac{x^{2}}{9}-\frac{4 y^{2}}{81}=1\)

(ix) Let the required equation of hyperbola be
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Length of transverse axis = 2a
Given, length of transverse axis = 8
⇒ 2a = 8
⇒ a = 4
⇒ a² = 16
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a²e² = 25
Now, b² = a² (e² – 1)
⇒ b² = a² e² – a²
⇒ b² = 25 – 16 = 9
The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Question 6.
Find the equation of the tangent to the hyperbola.
(i) 3x² – y² = 4 at the point (2, 2√2).
(ii) 3x² – y² = 12 at the point (4, 6)
(iii) \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\) at the point whose eccentric angle is \(\frac{\pi}{3}\).
(iv) \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\) at the point in a first quadrant whose ordinate is 3.
(v) 9x² – 16y² = 144 at the point L of the latus rectum in the first quadrant.
Solution:

Question 7.
Show that the line 3x – 4y + 10 = 0 is a tangent to the hyperbola x² – 4y² = 20. Also, find the point of contact.
Solution:
Given equation of the hyperbola is x² – 4y² = 20
\(\frac{x^{2}}{20}-\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 20 and b² = 5
Given equation of line is 3x – 4y + 10 = 0.
y = \(\frac{3 x}{4}+\frac{5}{2}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{3}{4}\) and c = \(\frac{5}{2}\)
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have

Question 8.
If the line 3x – 4y = k touches the hyperbola \(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\), then find the value of k.
Solution:
Given equation of the hyperbola is
\(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\)
\(\frac{x^{2}}{5}-\frac{y^{2}}{\frac{5}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 5, b² = \(\frac{5}{4}\)
Given equation of line is 3x – 4y = k
y = \(\frac{3}{4} x-\frac{\mathrm{k}}{4}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{3}{4}\), c = \(-\frac{\mathrm{k}}{4}\)
For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have
c² = a² m² – b²
⇒ \(\left(\frac{-\mathrm{k}}{4}\right)^{2}=5\left(\frac{3}{4}\right)^{2}-\frac{5}{4}\)
⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(9-4)\)
⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(5)\)
⇒ k² = 25
⇒ k = ±5

Alternate method:
Given equation of the hyperbola is
\(\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1\) …….(i)
Given equation of the line is 3x – 4y = k
y = \(\frac{3 x-\mathrm{k}}{4}\)
Substituting this value ofy in (i), we get
\(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{3 x-\mathrm{k}}{4}\right)^{2}=1\)
⇒ \(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{9 x^{2}-6 k x+k^{2}}{16}\right)=1\)
⇒ 4x² – (9x² – 6kx + k²) = 20
⇒ 4x² – 9x² + 6kx – k² = 20
⇒ -5x² + 6kx – k² = 20
⇒ 5x² – 6kx + (k² + 20) = 0 …..(ii)
Since, the given line touches the given hyperbola.
The quadratic equation (ii) in x has equal roots.
(-6k)² – 4(5)(k² + 20) = 0
⇒ 36k² – 20k² – 400 = 0
⇒ 16k² = 400
⇒ k² = 25
⇒ k = ±5

Question 9.
Find the equations of the tangents to the hyperbola \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\) making equal intercepts on the co-ordinate axes.
Solution:
Given equation of the hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 25 and b² = 9
Since the tangents make equal intercepts on the co-ordinate axes,
∴ m = -1
Equations of tangents to the hyperbola \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
⇒ y = -x ± \(\sqrt{25(-1)^{2}-9}\)
⇒ y = -x ± √16
⇒ x + y = ±4

Question 10.
Find the equations of the tangents to the hyperbola 5x² – 4y² = 20 which are parallel to the line 3x + 2y + 12 = 0.
Solution:
Given equation of the hyperbola is 5x² – 4y² = 20
\(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 4 and b² = 5
Slope of the line 3x + 2y + 12 = 0 is \(-\frac{3}{2}\)
Since the given line is parallel to the tangents,
Slope of the required tangents (m) = \(-\frac{3}{2}\)
Equations of tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1