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Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.4

Question 1.
Two dice are thrown simultaneously, if at least one of the dice shows a number 5, what is the probability that sum of the numbers on two dice is 9?
Solution:
When two dice are thrown simultaneously, 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 A be the event that at least one die shows number 5.
∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ n(A) = 11
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{11}{36}\)
Let B be the event that sum of the numbers on two dice is 9.
∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)}
Also, A ∩ B = {(4, 5), (5, 4)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{2}{36}\)
∴ Probability of sum of numbers on two dice is 9, given that one dice shows number 5, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{2}{36}}{\frac{11}{36}}=\frac{2}{11}\)

Question 2.
A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Solution:
When two dice are thrown simultaneously, 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 A be the event that sum of the numbers is an even number.
∴ A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
∴ n(A) = 18
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{18}{36}\)
Let B be the event that sum of outcomes is a perfect square.
∴ B = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
Also, A n B= {(1, 3), (2, 2), (3, 1)}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}\)
∴ Probability of sum of the numbers is a perfect square, given that sum of numbers is an even number, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{3}{36}}{\frac{18}{36}}=\frac{3}{18}=\frac{1}{6}\)

Question 3.
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Solution:
Two tickets can be drawn from 11 tickets with replacement in 11 × 11 = 121 ways.
∴ n(S) = 121
Let A be the event that the sum of two numbers is even.
The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn.
There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11.
∴ n(A) = 6 × 6 + 5 × 5
= 36 + 25
= 61
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{61}{121}\)
Let B be the event that the numbers tickets drawn are odd
∴ n(B) = 6 × 6 = 36
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
Since 6 odd numbers are common between A and B.
∴ n(A ∩ B) = 6 × 6 = 36
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
∴ Probability of both the numbers are odd, given that sum is even, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{36}{121}}{\frac{61}{121}}=\frac{36}{61}\)

Question 4.
A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club card is drawn.
B: an ace card is drawn.
Determine whether events A and B are independent or not.
Solution:
One card can be drawn out of 52 cards in ⁵²C₁ ways.
∴ n(S) = ⁵²C₁
Let A be the event that a club card is drawn.
1 club card out of 13 club cards can be drawn in ¹³C₁ ways.
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Let B be the event that an ace card is drawn.
An ace card out of 4 aces can be drawn in ⁴C₁ ways.
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Since 1 card is common between A and B
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{1}{52}\) …….(i)
∴ P(A) × P(B) = \(\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}} \times \frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13 \times 4}{52 \times 52}=\frac{1}{52}\) …….(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Question 5.
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that,
(i) problem is solved?
(ii) problem is not solved?
(iii) exactly two students solve the problem?
Solution:
Let A be the event that student A can solve the problem.
B be the event that student B can solve the problem.
C be the event that student C can solve problem.
∴ P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\), P(C) = \(\frac{1}{5}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
P(B’) = 1 – P(B) = 1 – \(\frac{1}{4}\) = \(\frac{5}{4}\)
P(C’) = 1 – P(C) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Since A, B, C are independent events
∴ A’, B’, C’ are also independent events
(i) Let X be the event that problem is solved.
Problem can be solved if at least one of the three students solves the problem.
P(X) = P(at least one student solves the problem)
= 1 – P(no student solved problem)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)

(ii) Let Y be the event that problem is not solved
∴ P(Y) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
= \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= \(\frac{2}{5}\)

(iii) Let Z be the event that exactly two students solve the problem.
∴ P(Z) = P(A ∩ B ∩ C’) ∪ P(A ∩ B’ ∩ C) ∪ P(A’ ∩ B ∩ C)
= P(A) . P(B) . P(C’) + P(A) . P(B’) . P(C) + P(A’) . P(B) . P(C)
= \(\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right)\)
= \(\frac{4}{60}+\frac{3}{60}+\frac{2}{60}\)
= \(\frac{3}{20}\)

Question 6.
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Solution:
Let A be the event that man will be alive at 60.
∴ P(A) = 0.83
Let B be the event that a woman will be alive at 55.
∴ P(B) = 0.97
A ∩ B = Event that both will be alive.
Also, A and B are independent events
∴ P(both man and his wife will be alive) = P(A ∩ B)
= P(A) . P(B)
= 0.83 × 0.97
= 0.8051

Question 7.
In an examination, 30% of the students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subjects I and subject II. A student is selected at random, what is the probability that the student
(i) has failed in the subject I, if it is known that he is failed in subject II?
(ii) has failed in at least one subject?
(iii) has failed in exactly one subject?
Solution:
Let A be the event that the student failed in Subject I
B be the event that the student failed in Subject II
Then P(A) = 30% = \(\frac{30}{100}\)
P(B) = 20% = \(\frac{20}{100}\)
and P(A ∩ B) = 10% = \(\frac{10}{100}\)
(i) P (student failed in Subject I, given that he has failed in Subject II) = P(A/B)
\(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\left(\frac{10}{100}\right)}{\left(\frac{20}{100}\right)}=\frac{10}{20}=\frac{1}{2}\)

(ii) P(student failed in at least one subject) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-\frac{10}{100}\)
= 0.40

(iii) P(student failed in exactly one subject) = P(A) + P(B) – 2P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-2\left(\frac{10}{100}\right)\)
= 0.30

Question 8.
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Solution:
A be the event that first gun hits the target
B be the event that second gun hits the target
C be the event that third gun hits the target
P(A) = 0.5, P(B) = 0.6, P(C) = 0.8
∴ P(A’) = 1 – P(A) = 1 – 0.5 = 0.5
∴ P(B’) = 1 – P(B) = 1 – 0.6 = 0.4
∴ P(C’) = 1 – P(C) = 1 – 0.8 = 0.2
Now A, B, C are independent events
∴ A’, B’, C are also independent events.
∴ P (at least one hit is registered)
= 1 – P(no hit is registered)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – (0.5) (0.4) (0.2)
= 1 – 0.04
= 0.96

Question 9.
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that
(i) first is white and second is black?
(ii) one is white and the other is black?
Solution:
Total number of balls = 10 + 15 = 25
Let S be an event that two balls are drawn at random without replacement in succession
∴ n(S) = ²⁵C₁ × ²⁴C₁ = 25 × 24
(i) Let A be the event that the first ball is white and the second is black.
First white ball can be drawn from 10 white balls in ¹⁰C₁ ways
and second black ball can be drawn from 15 black balls in ¹⁵C₁ ways.
∴ n(A) = ¹⁰C₁ × ¹⁵C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{10} \mathrm{C}_{1} \times{ }^{15} \mathrm{C}_{1}}{25 \times 24}=\frac{10 \times 15}{25 \times 24}=\frac{1}{4}\)

Question 10.
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 ball is black?
Solution:
Total number of balls in the urn = 4 + 5 + 6 = 15
Two balls can be drawn without replacement in ¹⁵C₂ = \(\frac{15 \times 14}{1 \times 2}\) = 105 ways
∴ n(S) = 105
Let A be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in ⁴C₁ = 4 ways
and 1 non-black ball can be drawn out of remaining 11 non-black balls in ¹¹C₁ = 11 ways
∴ 1 black and 1 non black ball can be drawn in 4 × 11 = 44 ways
Also, 2 black balls can be drawn from 4 black balls in ⁴C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6 ways
∴ n(A) = 44 + 6 = 50
∴ Required probability = P(A) = \(\frac{n(A)}{n(S)}=\frac{50}{105}\) = \(\frac{10}{21}\)

Alternate Solution:
Total number of balls = 15
Required probability = 1 – P(neither of two balls is black)
Balls are drawn without replacement
Probability of first non-black ball drawn = \(\frac{11}{15}\)
Probability of second non-black ball drawn = \(\frac{10}{14}\)
Probability of neither of two balls is black = \(\frac{11}{15} \times \frac{10}{14}=\frac{11}{21}\)
Required probability = 1 – \(\frac{11}{21}\) = \(\frac{10}{21}\)

Question 11.
Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Solution:
Total number of balls in the urn = 5 + 3 + 7 = 15
Out of these 12 are non-blue balls.
Two balls can be drawn from 15 balls without replacement in ¹⁵C₂
= \(\frac{15 \times 14}{1 \times 2}\)
= 105 ways.
∴ n(S) = 105
Let A be the event that at least one ball is blue,
i.e., 1 blue and other non-blue or both are blue.
∴ n(A) = ³C₁ × ¹²C₁ + ³C₂
= 3 × 12 + 3
= 36 + 3
= 39
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{39}{105}=\frac{13}{35}\)

Alternate solution:
Total number of balls in the urn = 15
Required probability = 1 – P(neither of two balls is blue)
Balls are drawn one by one without replacement.
Probability of first non-blue ball drawn = \(\frac{12}{15}\)
Probability of second non-blue ball drawn = \(\frac{11}{14}\)
Probability of neither of two ball is blue = \(\frac{12}{15} \times \frac{11}{14}=\frac{22}{35}\)
∴ Required probability = 1 – \(\frac{22}{35}\) = \(\frac{13}{35}\)

Question 12.
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball ¡s drawn from each bag, what is the Probability that two balls are of the same colour?
Solution:
Let A be the event that a blue ball is drawn from each bag.
Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is
P(A) = \(\frac{4}{9} \times \frac{3}{10}\)
Let B be the event that a green ball is drawn from each bag.
Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is
P(B) = \(\frac{5}{9} \times \frac{7}{10}\)
Since both, the events are mutually exclusive and exhaustive events
∴ P(that both the balls are of the same colour) = P(both are of blue colour) or P(both are of green colour)
= P(A) + P(B)
= \(\frac{4}{9} \times \frac{3}{10}+\frac{5}{9} \times \frac{7}{10}\)
= \(\frac{12}{90}+\frac{35}{90}\)
= \(\frac{47}{90}\)

Question 13.
Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?
Solution:
Two cards are drawn from a pack of 52 cards with replacement.
∴ n(S) = 52 × 52
Let A be the event that two cards drawn are face cards.
First card from 12 face cards is drawn with replacement in ¹²C₁ = 12 ways
and second face card is drawn from 12 face card in ¹²C₁ = 12 ways after replacement.
∴ n(A) = 12 × 12
∴ P(that both the cards drawn are face cards) = P(A)
= \(\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.3

Question 1.
Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?
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 A be the event that sum of numbers on two dice is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}\)
Let B be the event that number on second die is greater than or equal to number on first die.
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)}
∴ n(B) = 21
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{21}{36}\)
Now, A ∩ B = {(1, 4), (2, 3)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{36}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{4}{36}+\frac{21}{36}-\frac{2}{36}\)
= \(\frac{23}{36}\)

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 red or face card?
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that a red card is drawn Red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{26}{52}\)
Let B be the event that a black card is drawn
∴ Black card can be drawn in ²⁶C₁ = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= \(\frac{52}{52}\)
= 1

(ii) Let A be the event that a red card is drawn
∴ red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Let B be the event that a face card is drawn There are 12 face cards in the pack of 52 cards
∴ 1 face card can be drawn in ¹²C₁ = 12 ways
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 red face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{52}\)
∴ Required probability = P(A ∪ B)
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.
Two cards are drawn from a pack of 52 cards. What is the probability that,
(i) both the cards are of the same colour?
(ii) both the cards are either black or queens?
Solution:
Two cards can be drawn from 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that both cards are red.
∴ 2 red cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways
∴ n(B) = ²⁶C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{25}{102}+\frac{25}{102}\)
= \(\frac{25}{51}\)

(ii) Let A be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are queens.
There are 4 queens in a pack of 52 cards
∴ 2 queen cards can be drawn in ⁴C₂ ways.
∴ n(B) = ⁴C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
There are two black queen cards.
∴ n(A ∩ B) = ²C₂ = 1

Question 4.
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is 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 30?
Solution:
Out of the 50 tickets, a ticket can be drawn in ⁵⁰C₁ = 50 ways.
∴ n(S) = 50
(i) Let A be the event that the number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49}
∴ n(A) = 7
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{50}\)
Let B be the event that the number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{50}\)
Now, A ∩ B = {4, 16, 36}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{50}\)
Required probability = P (A u B)
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{7}{50}+\frac{12}{50}-\frac{3}{50}\)
= \(\frac{8}{25}\)

(ii) Let A be the event that 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}
∴ n(A) = 15
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{15}{50}\)
Let B be the event that the number is greater than 30.
∴ B = {31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}
∴ n(B) = 20
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{20}{50}\)
Now, A ∩ B = {31, 37, 41, 43, 47}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{50}+\frac{20}{50}-\frac{5}{50}\)
= \(\frac{15+20-5}{50}\)
= \(\frac{3}{5}\)

Question 5.
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random
(i) passed at least one examination.
(ii) passed in exactly one examination.
(iii) failed in both examinations.
Solution:
Out of hundred students 1 student can be selected in ¹⁰⁰C₁ = 100 ways.
∴ n(S) = 100
Let A be the event that the student passed in the first examination.
Let B be the event that student passed in second examination.
∴ n(A) = 60, n(B) = 50 and n(A ∩ B) = 30

(i) P(student passed in at least one examination) = P(A ∪ B)
= P(A) + P(B) – P (A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-\frac{3}{10}\)
= \(\frac{4}{5}\)

(ii) P(student passed in exactly one examination) = P(A) + P(B) – 2.P(A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-2\left(\frac{3}{10}\right)\)
= \(\frac{1}{2}\)

(iii) P(student failed in both examinations) = P(A’ ∩ B’)
= P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)

Question 6.
If P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\). Find the values of the following probabilities.
(i) P(A ∩ B)
(ii) P(A ∩ B’)
(iii) P(A’ ∩ B)
(iv) P(A’ ∪ B’)
(v) P(A’ ∩ B’)
Solution:
Here, P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\)
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{1}{4}+\frac{2}{5}-\frac{1}{2}\)
= \(\frac{3}{20}\)

(ii) P(A’ ∩ B’) = P(A) – P(A ∩ B)
= \(\frac{1}{4}-\frac{3}{20}\)
= \(\frac{1}{10}\)

(iii) P(A’ ∩ B) = P(B) – P(A ∩ B)
= \(\frac{2}{5}-\frac{3}{20}\)
= \(\frac{1}{4}\)

(iv) P(A’ ∪ B’) = P(A ∩ B)’ …..[De Morgan’s law]
= 1 – P(A ∩ B)
= 1 – \(\frac{3}{20}\)
= \(\frac{17}{20}\)

(v) P(A’ ∩ B’) = P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)

Question 7.
A computer software company is bidding for computer programs A and B. The probability that the company will get software A is \(\frac{3}{5}\), the probability that the company will get software B is \(\frac{1}{3}\) and the probability that company will get both A and B is \(\frac{1}{8}\). What is the probability that the company will get at least one software?
Solution:
Let A be the event that the company will get software A.
∴ P(A) = \(\frac{3}{5}\)
Let B be the event that company will get software B.
∴ P(B) = \(\frac{1}{3}\)
Also, P(A ∩ B) = \(\frac{1}{8}\)
∴ P(the company will get at least one software) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{5}+\frac{1}{3}-\frac{1}{8}\)
= \(\frac{72+40-15}{120}\)
= \(\frac{97}{120}\)

Question 8.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let A be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in ¹³C₁ ways
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{52}=\frac{13}{52}\)
Let B be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in ⁴C₁ ways
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{52}=\frac{4}{52}\)
There is one queen card out of 4 which is also a heart card
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{52}=\frac{1}{52}\)
∴ P(card is a heart or a queen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\)
= \(\frac{13+4-1}{52}\)
= \(\frac{16}{52}\)
∴ P(A ∪ B) = \(\frac{4}{13}\)

Question 9.
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in ⁷C₄
= \(\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}\)
= 35 ways.
∴ n(S) = 35
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in ³C₃ ways while a girl can be selected in ⁴C₁ ways.
∴ n(A) = ³C₃ × ⁴C₁ = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{35}\)
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in ⁴C3 ways and a boy can be selected in ³C₁ ways.
∴ n(B) = ⁴C₃ × ³C₁ = 12
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{12}{35}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
= P(A) + P(B)
= \(\frac{4}{35}+\frac{12}{35}\)
= \(\frac{16}{35}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.2

Question 1.
A fair die is thrown two times. Find the chance that
(i) product of the numbers on the upper face is 12.
(ii) sum of the numbers on the upper face is 10.
(iii) sum of the numbers on the upper face is at least 10.
(iv) sum of the numbers on the upper face is 4.
(v) the first throw gives an odd number and the second throw gives a multiple of 3.
(vi) both the times die to show the same number (doublet).
Solution:
If a fair die is thrown twice, 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
(i) Let A be the event that the product of the numbers on uppermost face is 12.
∴ A = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(A) = 4
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{4}{36}=\frac{1}{9}\)

(ii) Let B be the event that sum of the numbers on uppermost face is 10.
∴ B = {(4, 6), (5, 5), (6, 4)}
∴ n(B) = 3
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

(iii) Let C be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
∴ C = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), ( 6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(iv) Let D be the event that sum of the numbers on uppermost face is 4.
∴ D = {(1, 3), (2, 2), (3, 1)}
∴ n(D) = 3
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}=\frac{1}{12}\)

(v) Let E be the event that 1st throw gives an odd number and 2nd throw gives multiple of 3.
∴ E = {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6)}
∴ n(E) = 6
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(vi) Let F be the event that both times die shows same number.
∴ F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(F) = 6
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

Question 2.
Two cards are drawn from a pack of 52 cards. Find the probability that
(i) both are black.
(ii) both are diamonds.
(iii) both are ace cards.
(iv) both are face cards.
(v) one is a spade and the other is a non-spade.
(vi) both are from the same suit.
(vii) both are from the same denomination.
Solution:
Two cards can be drawn from a pack of 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
(i) Let A be the event that both the cards drawn are black.
The pack of 52 cards contains 26 black cards.
∴ 2 cards can be drawn from them in ²⁶C₂ ways
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(ii) Let B be the event that both the cards drawn are diamond.
There are 13 diamond cards in a pack of 52 cards.
∴ 2 diamond cards can be drawn from 13 diamond cards in ¹³C₂ ways
∴ n(B) = ¹³C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iii) Let C be the event that both the cards drawn are aces.
In a pack of 52 cards, there are 4 ace cards.
∴ 2 ace cards can be drawn from 4 ace cards in ⁴C₂ ways
∴ n(C) = ⁴C₂
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iv) Let D be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
∴ 2 face cards can be drawn from 12 face cards in ¹²C₂ ways.
∴ n(D) = ¹²C₂
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(v) Let E be the event that out of the two cards drawn one is a spade and other is non-spade.
There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
∴ One spade card can be drawn from 13 spade cards in ¹³C₁ ways and one non-spade card can be drawn from 39 non-spade cards in 39C1 ways.
∴ n(E) = ¹³C₁ . ³⁹C₁
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \cdot{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}\)

(vi) Let F be the event that 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 a suit in ¹³C₂ ways.
A suit can be selected in 4 ways.
∴ n(F) = ¹³C₂ × 4
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(vii) Let G be the event that both the cards drawn are of same denominations.
A pack of cards has 13 denominations and 4 different cards for each denomination
∴ n(G) = 13 × ⁴C₂
∴ P(G) = \(\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}=\frac{13 \times{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

Question 3.
Four cards are drawn from a pack of 52 cards. Find the probability that
(i) 3 are Kings and 1 is Jack.
(ii) all the cards are from different suits.
(iii) at least one heart.
(iv) all cards are club and one of them is a jack.
Solution:
4 cards can be drawn out of 52 cards in ⁵²C₄ ways.
∴ n(S) = ⁵²C₄
(i) Let A be the event that out of the four cards drawn, 3 are kings and 1 is a jack.
There are 4 kings and 4 jacks in a pack of 52 cards.
∴ 3 kings can be drawn from 4 kings in ⁴C₃ ways.
Similarly, 1 jack can be drawn out of 4 jacks in ⁴C₁ ways.
∴ Total number of ways in which 3 kings and 1 jack can be drawn is ⁴C₃ × ⁴C₁
∴ n(A) = ⁴C₃ × ⁴C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(ii) Let B be the event that all the cards drawn are of different suits.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ A card can be drawn from each suit in ¹³C₁ ways.
∴ 4 cards can be drawn from 4 different suits in ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ ways.
∴ n(B) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(iii) Let C be the event that out of the four cards drawn at least one is a heart.
∴ C’ is the event that all 4 cards drawn are non-heart cards.
In a pack of 52 cards, there are 39 non-heart cards.
∴ 4 non-heart cards can be drawn in ³⁹C₄ ways.
∴ n(C’) = ³⁹C₄
∴ P(C’) = \(\frac{\mathrm{n}\left(\mathrm{C}^{\eta}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)
∴ P(C) = 1 – P(C’) = 1 – \(\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)

(iv) Let D be the event that all the 4 cards drawn are clubs and one of them is a jack.
In a pack of 52 cards, there are 13 club cards having 1 jack card.
∴ 1 jack can be drawn in ¹C₁ way and the other 3 cards can be drawn from remaining 12 club cards in ¹²C₃ ways.
∴ n(D) = ¹²C₃ × ¹C₁
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{3} \times{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

Question 4.
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is 1/3. The probability of yellow is 1/5.
(i) What is the probability of blackball?
(ii) How many balls are green, black, and yellow?
Solution:
(i) The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow (Y)
∴ P(G) = \(\frac{1}{3}\) and P(Y) = \(\frac{1}{5}\)
If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
∴ P(G) + P(B) + P(Y) = 1
∴ \(\frac{1}{3}\) + P(B) + \(\frac{1}{5}\) = 1
∴ P(B) + \(\frac{8}{15}\) = 1
∴ P(B) = 1 – \(\frac{8}{15}\) = \(\frac{7}{15}\)
∴ Probability of black ball is \(\frac{7}{15}\)

(ii) Total number of balls = 15 and
P(G) = \(\frac{1}{3}\), P(B) = \(\frac{7}{15}\), P(Y) = \(\frac{1}{5}\)
∴ number of green balls = \(\frac{1}{3}\) × 15 = 5
number of black balls = \(\frac{7}{15}\) × 15 = 7
and number of yellow balls = \(\frac{1}{5}\) × 15 = 3.

Question 5.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that the
(i) number on the ticket is divisible by 6?
(ii) number on the ticket is a perfect square?
(iii) number on the ticket is prime?
(iv) 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 ⁷⁵C₁ = 75 ways.
∴ n(S) = 75
(i) Let A be the event that number on ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12}{75}=\frac{4}{25}\)

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

(iii) Let C be the event that the 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}
∴ n(C) = 21
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{21}{75}=\frac{7}{25}\)

(iv) Let D be the event that number on 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) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{5}{75}=\frac{1}{15}\)

Question 6.
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains
(i) 3 boys and 2 girls
(ii) at least 3 boys.
Solution:
The group consists of 8 boys and 5 girls i.e., 8 + 5 = 13 persons.
A committee of 5 is to be formed from this group.
∴ 5 persons from 13 persons can be selected in ¹³C₅ ways
∴ n(S) = ¹³C₅
(i) Let A be the event that the committee contains 3 boys and 2 girls.
3 boys from 8 boys can be selected in ⁸C₃ ways and 2 girls from 5 girls can be selected in ⁵C₂ ways
∴ n(A) = ⁸C₃ . ⁵C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}}{{ }^{13} \mathrm{C}_{5}}\)

(ii) Let B be the event that the committee contains at least 3 boys (i.e., 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
∴ n(B) = ⁸C₃ . ⁵C₂ + ⁸C₄ . ⁵C₁ + ⁸C₅ . ⁵C₀
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}+{ }^{8} \mathrm{C}_{4} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{5} \cdot{ }^{5} \mathrm{C}_{0}}{{ }^{13} \mathrm{C}_{5}}\)

Question 7.
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in a socket. What is the probability that 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 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in ⁶C₃ ways.
∴ n(A’) = ⁶C₃
∴ P(A’) = \(\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}\)
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 8.
The letters of the word LOGARITHM are arranged at random. Find the probability that
(i) Vowels are always together.
(ii) Vowels are never together.
(iii) Exactly 4 letters between G and H
(iv) begins with O and ends with T
(v) Start with a vowel and ends with a consonant.
Solution:
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ⁹P₉ = 9! ways.
∴ n(S) = 9!
(i) Let A be the event that vowels are always together.
The word LOGARITHM consists of 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).
3 vowels can be arranged among themselves in = ³P₃ = 3! ways.
Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7) can be arranged in ⁷P₇ = 7! ways.
∴ n(A) = 3! × 7!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3 ! \times 7 !}{9 !}\)

(ii) Let B be the event that vowels are never together.
Consider the following arrangement
_C_C_C_C_C_C_
6 consonants create 7 gaps.
∴ 3 vowels can be arranged in 7 gaps in ⁷P₃ ways.
Also 6 consonants can be arranged among themselves in ⁶P₆ = 6! ways.
∴ n(B) = 6! × ⁷P₃
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6 ! \times{ }^{7} \mathrm{P}_{3}}{9 !}\)

(iii) Let C be the event that exactly 4 letters are arranged between G and H.
Consider the following arrangement
1 2 3 4 5 6 7 8 9
∴ Out of 9 places, G and H can occupy any one of following 4 positions in 4 ways.
1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th
Now, G and H can be arranged among themselves in ²P₂ = 2! =2 ways.
Also, the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways.
∴ n(C) = 4 × 2 × 7! = 8 × 7! = 8!
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{8 !}{9 !}=\frac{8 !}{9 \times 8 !}=\frac{1}{9}\)

(iv) Let D be the event that word begins with O and ends with T.
Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways
∴ n(D) = 7! × 1 × 1 = 7!
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{7 !}{9 !}\)

(v) Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in ⁷P₇ = 7! ways
∴ n(E) = 3 × 6 × 7!
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{3 \times 6 \times 7 !}{9 !}\)

Question 9.
The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Solution:
The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T).
6 letters in which A repeats twice can be arranged among themselves in \(\frac{6 !}{2 !}\) ways.
∴ n(S) = \(\frac{6 !}{2 !}\)
Let A be the event that vowels are always together.
3 vowels (A, A, I) can be arranged among themselves in \(\frac{3 !}{2 !}\) ways.
Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in ⁴P₄ = 4! ways.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.1

Question 1.
State the sample space and n(S) for the following random experiments.
(i) A coin is tossed twice. If a second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.
Solution:
(i) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown.
Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
∴ S = {HH, TH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
∴ n(S) = 14

(ii) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
Let A be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
∴ A = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event B.
∴ B = {HTT, HTH, TTT, TTH}
The sample space S of the experiment is A ∪ B.
∴ S = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6, HTT, HTH, TTT, TTH}
∴ n(S) = 16

Question 2.
In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Solution:
The bag contains 3 balls out of which one is black (B), one is red (R) and the other one is green (G).
Two balls are drawn one after the other, with replacement, from the bag.
∴ the sample space S is given by
S = {BB, BR, BG, RB, RR, RG, GB, GR, GG}
∴ n(S) = 9

Question 3.
A coin and die are tossed. State sample space of following events.
(i) A: Getting a head and an even number.
(ii) B: Getting a prime number.
(iii) C: Getting a tail and perfect square.
Solution:
When a coin and a die are tossed the sample space S is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) A: getting a head and an even number
∴ A = {H2, H4, H6}

(ii) B: getting a prime number
∴ B = {H2, H3, H5, T2, T3, T5}

(iii) C: getting a tail and a perfect square.
∴ C = {T1, T4}

Question 4.
Find the total number of distinct possible outcomes n(S) for each of the following random experiments.
(i) From a box containing 25 lottery tickets and 3 tickets are drawn at random.
(ii) From a group of 4 boys and 3 girls, any two students are selected at random.
(iii) 5 balls are randomly placed into five cells, such that each cell will be occupied.
(iv) 6 students are arranged in a row for photographs.
Solution:
(i) Let S be the event that 3 tickets are drawn at random from 25 tickets
∴ 3 tickets can be selected in ²⁵C₃ ways
∴ n(S) = ²⁵C₃
= \(\frac{25 \times 24 \times 23}{3 \times 2 \times 1}\)
= 2300

(ii) There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in ⁷C₂ ways.
∴ n(S) = ⁷C₂
= \(\frac{7 \times 6}{2 \times 1}\)
= 21

(iii) 5 balls have to be placed in 5 cells in such a way that each cell is occupied.
∴ The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
∴ a total number of ways of filling 5 cells such that each cell is occupied = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(iv) Six students can be arranged in a row for a photograph in ⁶P₆ = 6! ways.
∴ n(S) = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question 5.
Two dice are thrown. Write favourable outcomes for the following events.
(i) P: The sum of the numbers on two dice is divisible by 3 or 4.
(ii) Q: sum of the numbers on two dice is 7.
(iii) R: sum of the numbers on two dice is a prime number.
Also, check whether
(a) events P and Q are mutually exclusive and exhaustive.
(b) events Q and R are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, all possible outcomes are
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)}
(i) P: sum of the numbers on two dice is divisible by 3 or 4.
∴ P = {(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)}

(ii) Q: sum of the numbers on two dice is 7.
∴ Q = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(iii) R: sum of the numbers on two dice is a prime number.
∴ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
(a) P and Q are mutually exclusive events as P ∩ Q = φ and
P ∪ Q = {(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
∴ P and Q are not exhaustive events as P ∪ Q ≠ S.

(b) Q ∩ R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ Q ∩ R ≠ φ
Also, Q ∪ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} ≠ S
∴ Q and R are neither mutually exclusive nor exhaustive events.

Question 6.
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits
(i) is not taken into account.
(ii) is taken into account.
Solution:
(i) If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in ⁵²C₁ = 52 ways.
∴ n(S) = 52

(ii) If consideration of suits is taken into account, then one card can be drawn from each suit in ¹³C₁ × ⁴C₁
= 13 × 4
= 52 ways.
∴ n(S) = 52

Question 7.
Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box II contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
∴ the sample space is
S = {(H, R11), (H, R12), (H, R13), (H, B11), (H, B12), (T, R21), (T, R22), (T, B21), (T, B22), (T, B23), (T, B24)}

Question 8.
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn
(i) with replacement
(ii) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8.
Two cards are to be drawn from this bag.
(i) 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)}

(ii) 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)}

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Miscellaneous Exercise 6

Question 1.
Find the value of r if ⁵⁶C_r+6 : ⁵⁴P_r-1 = 30800 : 1
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R
The number of words starting with A = 5!
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
The number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
Find the number of ways of distributing n balls in n cells. What will be the number of ways if each cell must be occupied?
Solution:
There are n balls and n cells
(i) Every ball can be put in any of the n cells.
Number of distributions = n × n × …… × n = (n)ⁿ
(ii) For filling the first cell, n balls are available.
The first cell is filled in n ways.
The second cell is filled in (n – 1) ways
The third cell is filled in (n – 2) ways and so on.
the nth cell is tilled in one way.
Required number = n(n – 1)(n – 2) …… 1 = n!

Question 5.
Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Solution:
Taking CST as the first station and Thane as 20th,
Let us name CST as A₀ next station as A₁ and so on, Thane is A₂₀
From station A₀, 20 different journeys are possible
From station A₁, 20 different journeys are possible.
From station A₂₀, 20 different journeys are possible.
Total number of different tickets of different journeys = 21 × 20 = 420

Question 6.
English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 7.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360
∴ 360 numbers that exceed one million can be formed with the digits 3, 2, 0, 4, 3, 2, 3.

Question 8.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:

Required number = ²⁶C₆ + ²⁶C₁₀

Question 9.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:

Required Number = 5 + 10 = 15

Question 10.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so. Solution:
Required number = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃

Question 11.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
∴ Required number = 128 – 1 = 127

Question 12.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:

Question 13.
Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Solution:
Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then A ∪ B ∪ C represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number = Total number of distributions – n(A ∪ B ∪ C) …..(i)
n(A ∪ B ∪ C) represent the number of undesirable cases
Total number of distributions = 3 × 3 × 3 × 3 × 3 = 3⁵ = 243 …….(ii)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C) …..(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
∴ n(A) + n(B) + n(C) = 3 × 2⁵ ……(iv)
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 3 × 1⁵ ……(v)
∴ n(A ∩ B ∩ C) = 0 …….(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
n(A ∪ B ∪ C) = 3 × 2⁵ – 3 × 1⁵
= 96 – 3
= 93
Substitute n(A ∪ B ∪ C) and from (ii) to (i), we get
Required number = 243 – 93 = 150

Question 14.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes the case in which all 12 lamps are OFF.
∴ Required Number = 2¹² – 1 = 4095

Question 15.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0

∴ Required number = 3 × 4 × 4 = 48

Question 16.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
Let the telephone number be 45abcd

∴ Required number = ⁸P₄ = 1680

Question 17.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 question.
Number of outcomes = 2⁶
This number includes the case when the student solves NONE of the question.
Required number = 2⁶ – 1
= 64 – 1
= 63

Question 18.
Find the number of ways of dividing 20 objects in three groups of sizes 8, 7 and 5.
Solution:
Select 8 object out of 20 in ²⁰C₈ ways
Select 7 object from remaining 12 in ¹²C₇ ways and 5 objects from remaining 5 in ⁵C₅ ways
Required number is = ²⁰C₈ × ¹²C₇ × ⁵C₅

Question 19.
There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since, there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number = ¹²C₆ = 924

Question 20.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
Solution:
We need 2 lines from each set.
Required number = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.7 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.7

Question 1.
Find n if ⁿC₈ = ⁿC₁₂
Solution:
ⁿC₈ = ⁿC₁₂
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

Question 2.
Find n if ²³C_3n = ²³C_2n+3
Solution:
²³C_3n = ²³C_2n+3
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

Question 3.
Find n if ²¹C_6n = \({ }^{21} C_{n^{2}+5}\)
Solution:
²¹C_6n = \({ }^{21} C_{n^{2}+5}\)
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 6n = n² + 5 or 6n = 21 – (n² + 5)
∴ n² – 6n + 5 = 0 or 6n = 21 – n² – 5
∴ n² – 6n + 5 = 0 or n² + 6n – 16 = 0
If n² – 6n + 5 = 0 then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then n² + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n² + 6n – 16 = 0 then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2

Question 4.
Find n if ²ⁿC_r-1 = ²ⁿC_r+1
Solution:
²ⁿC_r-1 = ²ⁿC_r+1
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Question 5.
Find n if ⁿC_n-2 = 15
Solution:
ⁿC_n-2 = 15
∴ ⁿC₂ = 15 …….[∵ ⁿC_r = ⁿC_n-r]
∴ \(\frac{n !}{(n-2) ! 2 !}\) = 15
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) ! 2 \times 1}\) = 15
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Comparing both sides, we get
∴ n = 6

Question 6.
Find x if ⁿP_r = x ⁿC_r
Solution:

Question 7.
Find r if ¹¹C₄ + ¹¹C₅ + ¹²C₆ + ¹³C₇ = ¹⁴C_r
Solution:

Question 8.
Find the value of \(\sum_{r=1}^{4}{ }^{21-r} C_{4}+{ }^{17} C_{5}\)
Solution:

Question 9.
Find the differences between the largest values in the following:
(i) ¹⁴C_r – ¹²C_r
Solution:
Greatest value of ¹⁴C_r
Here n = 14, which is even
Greatest value of ⁿC_r occurs at r = \(\frac{n}{2}\) if n is even

∴ Difference between the greatest values of ¹⁴C_r and ¹²C_r = 3432 – 924 = 2508

(ii) ¹³C_r – ⁸C_r
Solution:
Greatest value of ¹³C_r
Here n = 13, which is odd
Greatest value of ⁿC_r occurs at r = \(\frac{\mathrm{n}-1}{2}\) if n is odd

∴ Difference between the greatest values of ¹³C_r and ⁸C_r = 1716 – 70 = 1646

(iii) ¹⁵C_r – ¹¹C_r
Solution:
Greatest value of ¹⁵C_r
Here n = 15, which is odd
Greatest value of ⁿC_r occurs at r = \(\frac{\mathrm{n}-1}{2}\) if n is odd

Difference between the greatest values of ¹⁵C_r and ¹¹C_r = 6435 – 462 = 5973

Question 10.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more join the party = 10 + 5 + 1 = 16

Question 11.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consists of at least 3 women.

∴ Number of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 12.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in the majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women
Consider the following table:


∴ Number of committees = 14112 + 14700 + 6720 + 1260 + 81 = 36873
∴ At least 5 women are there in 36873 committees.

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= ¹⁸C₁₀ – 36873
= ¹⁸C₈ – 36873
= 43758 – 36873
= 6885

Question 13.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 14.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicketkeeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected = 45045 + 6006 = 51051

Question 15.
Five students are selected from 11. How many ways can these students be selected if
(i) two specified students are selected?
(ii) two specified students are not selected?
Solution:
5 students are to be selected from 11 students
(i) When 2 specified students are included
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = ⁹C₃
= \(\frac{9 !}{3 ! \times 6 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\)
= 84
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

(ii) When 2 specified students are not included then 5 students can be selected from the remaining (11 – 2) = 9 students
∴ Number of ways of selecting 5 students from 9 students = ⁹C₅
= \(\frac{9 !}{5 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}\)
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.6

Question 1.
Find the value of
(i) ¹⁵C₄
Solution:

(ii) ⁸⁰C₂
Solution:

(iii) ¹⁵C₄ + ¹⁵C₅
Solution:

(iv) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(i) ⁶P₂ = n ⁶C₂
Solution:

(ii) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(iii) ⁿC_n-3 = 84
Solution:
ⁿC_n-3 = 84
∴ \(\frac{n !}{(n-3) ![n-(n-3)] !}\) = 84
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}\) = 84
∴ n(n – 1) (n – 2) = 84 × 6
∴ n(n – 1) (n – 2) = 9 × 8 × 7
Comparing on both sides, we get
∴ n = 9

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10
Solution:

∴ \(\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}\)
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if.
(i) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(ii) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find r.
Solution:

∴ r! = 40320
∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
∴ r! = 8!
∴ r = 8

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 green balls can be selected from 5 green balls in ⁵C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected = ⁶C₃ × ⁴C₂
= \(\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}\)
= 20 × 6
= 120
∴ The team of 3 boys and 2 girls can be selected in 120 ways.

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
∴ 66 = ⁿC₂
∴ 66 = \(\frac{n !}{2 !(n-2) !}\)
∴ 66 × 2 = \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}\)
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides. .
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (∴ n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 10 points on a plane.
(i) No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
= ¹⁰C₂ – ⁴C₂ + 1
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 12 points on the plane
(i) When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
Out of 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ i.e., 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.5

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways
∴ 8 friends can sit around a table in 7! ways
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.
∴ 8 friends can sit around a table in 5040 ways.

Question 2.
A party has 20 participants and a host. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants be seated on either side of the host.
Host takes chair in 1 way.
These 2 persons can sit on either side of host in 2! ways
Once host occupies his chair, it is not circular permutation any more.
Remaining 18 people occupy their chairs in 18! ways.
∴ Total number of arrangement possible if two particular participants be seated on either side of the host = 2! × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are
(i) always together.
(ii) never together.
Solution:
(i) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in (23 – 1)! = 22! ways.
∴ The total number of arrangements if two specified delegates are always together = 22! × 2!

(ii) When 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Total number of arrangements if two specified delegates are never together = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !}\) × 21!
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in (15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e. clockwise and anticlockwise) coincide.

∴ Number of arrangements possible for not to have same neighbours = \(\frac{14 !}{2}\)

Question 5.
A committee of 20 members sits around a table. Find the number of arrangements that have the president and the vice president together.
Solution:
A committee of 20 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in (19 – 1)! = 18! ways.
∴ The total number of arrangements possible if President and Vice-president sit together = 18! × 2!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated
(i) between the two women.
(ii) between two men.
Solution:
(i) 5 men, 2 women, and a child sit around a table
When a child is seated between two women
∴ The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Also, these 3 are to be seated with 5 men, (i.e. a total of 6 units) which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements if the child is seated between two women = 5! × 2!

(ii) Two men out of 5 men can sit on either side of the child in 5P2 ways.
Let us take two men and a child as one unit.
Now these are to be arranged with the remaining 3 men and 2 women
i.e., a total of 6 events (3 + 2 + 1) is to be arranged around a round table which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements, if the child is seated between two men = ⁵P₂ × 5!

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
There are 8 gaps created by 8 men’s seats.
∴ 6 Women can be seated in 8 gaps in ⁸P₆ ways
∴ Total number of arrangements so that no two women are together = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:

Two women sit together and one woman sits separately.
Women sitting separately can be selected in 3 ways.
The other two women occupy two chairs in one way (as it is a circular arrangement).
They can be seated on those two chairs in 2 ways. Suppose two chairs are chairs 1 and 2 shown in the figure.
Then the third woman has only two options viz chairs 4 or 5.
∴ The third woman can be seated in 2 ways. 3 men are seated in 3! ways
∴ Required number = 3 × 2 × 2 × 3!
= 12 × 6
= 72

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required total number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side.
The other 13 persons can be arranged around the table in (13 – 1)! = 12!
13 people around a table create 13 gaps in which 2 people are to be seated
Number of arrangements of 2 people = ¹³P₂
∴ The total number of arrangements in which two specified persons not sitting side by side = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.4

Question 1.
Find the number of permutations of letters in each of the following words:
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
Solution:
(i) There are 5 letters in the word DIVYA which can be arranged in 5! Way = 120 ways

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ repeats 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9 !}{3 !}\)
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ repeats 3 times and ‘R’ repeats 2 times.
∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9 !}{3 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}\)
= 30240

(iv) There are 7 distinct letters in the word COMBINE which can be arranged among themselves in 7! = 5040 ways

Question 2.
You have 2 identical books on English, 3 identical books on Hindi and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on Mathematics are identical.
∴ Total number of arrangements = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260
∴ In 1260 distinct ways the books can be arranged on a shelf.

Question 3.
A coin is tossed 8 times. In how many ways can we obtain
(i) 4 heads and 4 tails?
(ii) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(i) We can obtain 4 heads and 4 tails in \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70 ways
∴ In 70 different ways we can obtain 4 heads and 4 tails.

(ii) When at least 6 heads are to be obtained
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways in which it can be obtained = \(\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}\)
= \(\frac{8 \times 7}{2}\) + 8 + 1
= 28 + 8 + 1
= 37
∴ In 37 different ways we can obtain at least 6 heads.

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = \(\frac{13 !}{5 ! 4 ! 4 !}\)

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ repeats 2 times, ‘A’ repeats 3 times, and ‘T’ repeats 2 times.
∴ Total number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels
i.e., ‘A’, ‘A’, ‘A’, ‘E’, T are to be kept together
Number of arrangements of these vowels = \(\frac{5 !}{3 !}\) ways.
Let us consider these vowels together as one unit.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Total number of arrangements = \(\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}\)

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have
(i) letters M and T never together?
(ii) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of arrangements is \(\frac{11 !}{4 ! 2 ! 2 !}\)
∴ \(\frac{11 !}{4 ! 2 ! 2 !}\) different words can be formed from the letters of the word MAHARASHTRA.
(i) Other than M and T. there are 9 letters in which A repeats 4 times, H repeats twice, R repeats twice
The number of arrangements of the a letter = \(\frac{9 !}{4 ! 2 ! 2 !}\)
These 9 letters create 10 gaps in which M and T are to be arranged
The number of arrangements of M and T = ¹⁰P₂
∴ Total number arrangement having M and T never together = \(\frac{9 ! \times{ }^{10} \mathrm{P}_{2}}{4 ! 2 ! 2 !}\)

(ii) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA i.e., A, A, A, A
Let us consider these 4 vowels as one unit, they themselves can be arranged in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters which can be done in 8! ways
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ \(\frac{8 !}{2 ! 2 !}\) different words can be formed if vowels are always together.

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
When ‘R’ is used thrice, ‘S’ is used twice and ‘T’ is used twice,
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of arrangements = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210
∴ 210 different words can be formed with the letter R is used thrice and letters S and T are used twice each.

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit. This unit with other 4 letters in which ‘M’ repeats twice, is to be arranged.
∴ Total number of arrangements when B is always next to A = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, T repeat two times each, ‘T’ repeats 3 times.
The arrangement starts and ends with ‘N’, 10 letters other than N can be arranged between two N, in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Total number of arrangements with the letter N at the beginning and at the end = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements have two R’s and two A’s not together?
Solution:
(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and R is repeated 2 times
∴ The number of arrangements = \(\frac{7 !}{2 ! 2 !}\) = 1260

(ii) A: set of words having 2A together
B: set of words having 2R together
Number of words having both A and both R not together
= 1260 – n(A ∪ B)
= 1260 – [n(a) + n(B) – n(A ∩ B)] ……(i)
n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged
∴ n(A) = \(\frac{6 !}{2 !}\) = 360
n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged
∴ n(B) = \(\frac{6 !}{2 !}\) = 360
n(A ∩ B) = number of ways in which (AA), (RR), N, G, E are to be arranged
∴ n(A ∩ B) = 5! = 120
Substituting n(A), n(B), n(A ∩ B) in (i), we get
Number of words having both A and both R not together
= 1260 – [360 + 360 – 120]
= 1260 – 600
= 660

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \(\frac{5 !}{2 !}\) × 2
= 5!
= 120

Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}\) × 2 = 60
Required number = 120 + 60 = 180
∴ 180 distinct 5 digit numbers can be formed using the digit 3, 2, 3, 2, 4, 5.

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits i.e. 3, 5, 7, 9.
They can be arranged at 4 odd places among themselves in 4! ways = 24 ways
3 even places of the number are occupied by even digits (i.e. 4, 6, 8).
∴ They can be arranged in 3! ways = 6 ways
∴ Total number of arrangements = 24 × 6 = 144
∴ 144 numbers can be formed so that odd digits always occupy the odd positions.

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 2?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Total number of arrangements = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
∴ 360 different 6-digit numbers can be formed.
For a number to be divisible by 2,
Last digits can be selected in 3 ways
Remaining 5 digit in which, 9 appears twice are arranged in \(\frac{5 !}{2 !}\) ways
∴ Total number of arrangements = \(\frac{5 !}{2 !}\) × 3 = 180
∴ 180 numbers are divisible by 2.

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N repeat twice.
Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}\)
= 180
∴ 180 different words can be formed with the letters of the word INDIAN.
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60 ways.
∴ 2N can be arranged in 1 way
∴ Total number of arrangements = 60 × 1 = 60 ways
∴ 60 words are such that two N’s are together.

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if
(i) the two O’s are never together.
(ii) consonants and vowels occupy alternate positions.
Solution:
(i) When the two O’s are never together:
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
∴ Two O’s in 6 gaps can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\) ways
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\) ways
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\) ways
= 3 × 5 ways
= 15 ways
∴ Total number of arrangements if the two O’s are never together = 120 × 15 = 1800

(ii) When consonants and vowels occupy alternate positions:
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places consonants occur and at even places vowels occur.
4 consonants can be arranged among themselves in 4! ways
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in \(\frac{3 !}{2 !}\) ways
∴ Required number of arrangements if the consonants and vowels occupy alternate positions = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.3

Question 1.
Find n if ⁿP₆ : ⁿP₃ = 120 : 1
Solution:
ⁿP₆ : ⁿP₃ = 120 : 1
∴ \(\frac{n !}{(n-6) !} \div \frac{n !}{(n-3) !}=\frac{120}{1}\)
∴ \(\frac{\mathrm{n} !}{(\mathrm{n}-6) !} \times \frac{(\mathrm{n}-3) !}{\mathrm{n} !}\) = 120
∴ \(\frac{n !}{(n-6) !} \times \frac{(n-3)(n-4)(n-5)(n-6) !}{n !}\) = 120
∴ (n – 3) (n – 4) (n – 5) = 120
∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4
Comparing on both sides, we get
n – 3 = 6
∴ n = 9

Question 2.
Find m and n if ^(m+n)P₂ = 56 and ^(m-n)P₂ = 12
Solution:
^m+nP₂ = 56
∴ \(\frac{(\mathrm{m}+\mathrm{n}) !}{(\mathrm{m}+\mathrm{n}-2) !}\) = 56
∴ \(\frac{(m+n)(m+n-1)(m+n-2) !}{(m+n-2) !}\) = 56
∴ (m + n) (m + n – 1) = 8 × 7
Comparing on both sides, we get
m + n = 8 …..(i)
Also ^m-nP₂ = 12
∴ \(\frac{(m-n) !}{(m-n-2) !}\) = 12
∴ \(\frac{(m-n)(m-n-1)(m-n-2) !}{(m-n-2) !}\) = 12
∴ (m – n) (m – n – 1) = 4 × 3
Comparing on both sides, we get
m – n = 4 …..(ii)
Adding (i) and (ii), we get
2m = 12
∴ m = 6
Substituting m = 6 in (ii), we get
6 – n = 4
∴ n = 2

Question 3.
Find r if ¹²P_r-2 : ¹¹P_r-1 = 3 : 14
Solution:

∴ (14 – r)(13 – r) = 8 × 7
Comparing on both sides, we get
14 – r = 8
∴ r = 6

Question 4.
Show that (n + 1) ⁿP_r = (n – r + 1) ⁽ⁿ⁺¹⁾P_r.
Solution:

Question 5.
How many 4 letter words can be formed using letters in the word MADHURI if
(i) letters can be repeated?
(ii) letters cannot be repeated?
Solution:
There are 7 letters in the word MADHURI.
(i) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
∴ 1st letter can be filled in 7 ways.
2nd letter can be filled in 7 ways.
3rd letter can be filled in 7 ways.
4th letter can be filled in 7 ways.
∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401
∴ 2401 four-lettered words can be formed when repetition of letters is allowed.

(ii) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is
⁷P₄ = \(\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}\) = 840
∴ 840 four-letter words can be formed when repetition of letters is not allowed.

Alternate method:
There are 7 letters in the word MADHURI.
(i) Since letters can be repeated
∴ In all places of a four-letter word, any one of seven letters M, A, D, H, U, R, I can appear.
∴ Using the Multiplication theorem, we get
Number of four-letter words with repetition of letters M, A, D, H, U, R, I = 7 × 7 × 7 × 7 = 2401

(ii) Since the letters cannot be repeated therefore 1st, 2nd, 3rd, 4th places can be filled in 7, 6, 5, 4 ways respectively
∴ Using the multiplication theorem, we get
Number of four-letter words, with no repetition of letters M, A, D, H, U, R, I = 7 × 6 × 5 × 4 = 840

Question 6.
Determine the number of arrangements of letters of the word ALGORITHM if
(i) vowels are always together.
(ii) no two vowels are together.
(iii) Consonants are at even positions
(iv) O is first and T is last.
Solution:
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
(i) When vowels are always together:
There are 3 vowels in the word ALGORITHM. (i.e, A, I, O)
Let us consider these 3 vowels as one unit.
This unit with 6 other letters is to be arranged.
∴ It becomes an arrangement of 7 things which can be done in ⁷P₇ i.e., 7! ways and 3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
∴ the total number of ways in which the word can be formed = 7! × 3!
= 5040 × 6
= 30240
∴ 30240 words can be formed if vowels are always together.

(ii) When no two vowels are together:
There are 6 consonants in the word ALGORITHM.
They can be arranged among themselves in ⁶P₆ i.e., 6! ways.
Let consonants be denoted by C.
_C_C_ C_C_C_C_
6 consonants create 7 gaps in which 3 vowels are to arranged.
∴ 3 vowels can be filled in ⁷P₃
= \(\frac{7 !}{(7-3) !}\)
= \(\frac{7 \times 6 \times 5 \times 4 !}{4 !}\)
= 210 ways
∴ total number of ways in which the word can be formed = 6! × 210
= 720 × 210
= 151200
∴ 151200 words can be formed if no two vowels are together.

(iii) When consonants are at even positions:
There are 4 even places and 6 consonants in the word ALGORITHM.
1st, 2nd, 3rd, 4th even places are filled in 6, 5, 4, 3 way respectively.
∴ The number of ways to fill four even places by consonants = 6 × 5 × 4 × 3 = 360
The remaining 5 letters (3 vowels and 2 consonants) can be arranged among themselves in ⁵P₅ i.e., 5! ways.
∴ Total number of ways the words can be formed
In which even places are occupied by consonants = 360 × 5!
= 360 × 120
= 43200
∴ 43200 words can be formed if even positions are occupied by consonants.

(iv) When beginning with O and ends with T:
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
7 letters other than O and T can be filled between O and T in ⁷P₇ i.e., 7! ways = 5040 ways.
∴ 5040 words beginning with O and ending with T can be formed.

Question 7.
In a group photograph, 6 teachers and principals are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Solution:
In 1st row middle seat is fixed for the principal.
Also 1st row, 6 teachers can be arranged among themselves in ⁶P₆ i.e., 6! ways.
In the 2nd row, 12 boys can be arranged among themselves in ¹²P₁₂ i.e., 12! ways.
13 gaps are created by 12 boys, in which 6 girls are to be arranged.
together which can be done in ¹³P₆ ways.
∴ total number of arrangements = 6! × 12! × ¹³P₆ …..[using Multiplications Principle]
= 6! × 12! × \(\frac{13 !}{(13-6) !}\)
= 6! × 12! × \(\frac{13 !}{7 !}\)
= \(\frac{6 ! \times 12 ! \times 13 !}{7 \times 6 !}\)
= \(\frac{12 ! 13 !}{7}\)

Question 8.
Find the number of ways letters of the word HISTORY can be arranged if
(i) Y and T are together
(ii) Y is next to T.
Solution:
There are 7 letters in the word HISTORY
(i) When ‘Y’ and ‘T’ are together.
Let us consider ‘ Y’ and ‘T’ as one unit
This unit with the other 5 letters is to be arranged.
∴ The number of arrangements of one unit and 5 letters = ⁶P₆ = 6!
Also, ‘Y’ and ‘T’ can be arranged among themselves in ²P₂ i.e., 2! ways.
∴ a total number of arrangements when Y and T are always together = 6! × 2!
= 720 × 2
= 1440
∴ 1440 words can be formed if Y and T are together.

(ii) When ‘Y’ is next to ‘T’
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 6 letters and one unit = ⁶P₆ = 6!
Also ‘Y’ has to be always next to ‘T’.
So they can be arranged in 1 way.
∴ total number of arrangements possible when Y is next to T = 6! × 1 = 720
∴ 720 words can be formed if Y is next to T.

Question 9.
Find the number of arrangements of the letters in the word BERMUDA so that consonants and vowels are in the same relative positions.
Solution:
There are 7 letters in the word “BERMUDA” out of which 3 are vowels and 4 are consonants.
If relative positions of consonants and vowels are not changed.
3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
4 consonants can be arranged among themselves in ⁴P₄ i.e., 4! ways.
∴ total no. of arrangements possible if relative positions of vowels and consonants are not changed = 3! × 4!
= 6 × 24
= 144

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
(i) digits can be repeated
(ii) digits cannot be repeated
Solution:
(i) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
∴ The unit’s place digit can be filled in 6 ways.
10’s place digit can be filled in 6 ways.
100’s place digit can be filled in 6 ways.
1000’s place digit can be filled in 6 ways.
∴ total number of numbers = 6 × 6 × 6 × 6 = 6⁴ = 1296
∴ 1296 four-digit numbers can be formed if repetition of digits is allowed.

(ii) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
∴ 4 different digits are to be arranged from 6 given digits which can be done in ⁶P₄
= \(\frac{6 !}{(6-4) !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360 ways
∴ 360 four-digit numbers can be formed, if repetition of digits is not allowed.

Question 11.
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that the resulting numbers are between 100 and 1000?
Solution:
A number between 100 and 1000 is a 3 digit number and is to be formed from the digits 0, 1, 2, 3, 4, 5, without repetition of digits.
∴ 100’s place digit must be a non-zero number which can be filled in 5 ways.
10’s place digits can be filled in 5 ways.
Unit’s place digit can be filled in 4 ways.
∴ total number of ways the number can be formed = 5 × 5 × 4 = 100
∴ 100 numbers between 100 and 1000 can be formed.

Question 12.
Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are
(i) divisible by 5?
(ii) not divisible by 5?
Solution:
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in ⁶P₆ i.e., 6! = 720 ways
(i) If the number is divisible by 5, then
The unit’s place digit must be 5, and hence unit’s place can be filled in 1 way
Other 5 digits can be arranged among themselves in ⁵P₅ i.e., 5! ways
∴ Total number of ways in which numbers divisible by 5 can be formed = 1 × 5! = 120

(ii) If the number is not divisible by 5, then
Unit’s place can be any digit from 3, 4, 6, 7, 8 which can be selected in 5 ways.
Other 5 digits can be arranged in ⁵P₅ i.e., 5! ways
∴ The total number of ways in which numbers not divisible by 5 can be formed = 5 × 5!
= 5 × 120
= 600

Question 13.
A code word is formed by two distinct English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.
Solution:
(i) There is a total of 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26 !}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways
Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in ⁹P₂
= \(\frac{9 !}{(9-2) !}\)
= \(\frac{9 \times 8 \times 7 !}{7 !}\)
= 72 ways
∴ Total number of a code words = 650 × 72 = 46800

(ii) There are total 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways
For a code word to end with an even integer, the digit in the unit’s place should be an even number between 1 to 9 which can be filled in 4 ways.
Also, 10’s place can be filled in 8 ways.
∴ Total number of codewords = 650 × 4 × 8 = 20800 ways
∴ 20800 codewords end with an even integer.

Question 14.
Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Solution:
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
∴ Each letter can be posted in 3 ways.
∴ Total number of ways in which 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243

Question 15.
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object
(i) always occurs
(ii) never occurs
Solution:
There are 11 distinct objects and 4 are to be taken at a time.
(i) The number of permutations of n distinct objects, taken r at a time, when one specified object will always occur is r × ^(n-1)P_(r-1)
Here, r = 4, n = 11
∴ The number of permutations of 4 out of 11 objects when a specified object occurs.
= 4 × ^(11-1)P_(4-1)
= 4 × ¹⁰P₃
= 4 × \(\frac{10 !}{(10-3) !}\)
= 4 × \(\frac{10 !}{7 !}\)
= 4 × \(\frac{10 \times 9 \times 8 \times 7 !}{7 !}\)
= 2880
∴ There are 2880 permutations of 11 distinct objects, taken 4 at a time, in which one specified object always occurs.

(ii) When one specified object does not occur then 4 things are to be arranged from the remaining 10 things, which can be done in ¹⁰P₄ ways
= 10 × 9 × 8 × 7 ways
= 5040 ways
∴ There are 5040 permutations of 11 distinct objects, taken 4 at a time, in which one specified object never occurs.