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

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)
= $$\frac{4}{35}+\frac{12}{35}$$
= $$\frac{16}{35}$$