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## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.2

Question 1.

(i) If P (A) = \(\frac{1}{2}\) P (B) = 0, then what can you say about P (A | B) ?

(ii) If A and B are events such that P(A | B) = P (B | A), then show that P(A) = P (B). (NCERT)

Answer:

(i) P(A) = \(\frac{1}{2}\); P(B) = 0

∴ P (A | B) = \(\frac{P(A \cap B)}{P(B)}\) is not defined [∵ P (B) = 0]

(ii) Given P (A | B) = P (B | A)

⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(B \cap A)}{P(A)}\)

⇒ P (A) = P (B) [∵ P (A ∩ B) = P (B ∩ A)]

Question 2.

If A and B are events such that P(A) = \(\frac{1}{2}\); P(B) = \(\frac{1}{3}\) and P(A ∩ B) = \(\frac{1}{4}\) then find

(i) P (A/B) (ii) P (B/A)

Answer:

Given P (A) = \(\frac{1}{2}\); P (B) = \(\frac{1}{3}\)

and P(A ∩ B) = \(\frac{1}{4}\)

(i) P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}\)

(ii) P(B/A) = \(\frac{P(B \cap A)}{P(A)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)

Question 3.

If the events A and B are such that P(A) = \(\frac{1}{4}\) , P(B/A) = \(\frac{2}{3}\), P(A/B) = \(\frac{1}{2}\) then find P(B).

Answer:

Given P(A) = \(\frac{1}{4}\) , P(B/A) = \(\frac{2}{3}\)

P(A/B) = \(\frac{1}{2}\)

Since P(B/A) = \(\frac{2}{3}\)

⇒ \(\frac{2}{3}=\frac{P(B \cap A)}{P(A)}\)

⇒ (A ∩ B) = \(\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}\)

Also, \(\frac{1}{2}\) = P(A/B) = \(\frac{P(A \cap B)}{P(B)}\)

⇒ \(\frac{1}{2}\) × P(B) = \(\frac{1}{6}\)

⇒ P(B) = \(\frac{1}{3}\)

Question 4.

If P(not A) = 0.7, P(B) = 0.7 and P(B|A) = 0.5, then find P(A|B) and P(A ∪ B)

Answer:

Given P(not A) = 0.7 ⇒ P(Ā) = 0.7

⇒ P(A) = 1 – P(Ā) = 1 – 0.7 = 0.3

and P(B) = 0.7; P(B/A) = 0.5

∴ 0.5 = \(\frac{P(B \cap A)}{P(A)}\)

⇒ P(B ∩ A) = 0.5 P(A) = 0.5 × 0.3 = 0.15

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.3 + 0.7 – 0.15 = 0.85

Thus, P(A|B) = \(\frac{P(A \cap B)}{P(B)}\)

= \(\frac{0.15}{0.7}=\frac{15}{70}=\frac{3}{14}\)

Question 5.

Compute P(A|B) when P(A) = \(\frac{1}{5}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{3}{5}\)

Answer:

Given P(A) = \(\frac{1}{5}\); P(B) = \(\frac{2}{5}\)

P(A ∪ B) = \(\frac{3}{5}\)

Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ \(\frac{3}{5}=\frac{1}{5}+\frac{2}{5}\) – P(A ∩ B)

⇒ P(A ∩ B) = 0

∴ P(A|B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0}{2 / 5}\) = 0

Question 6.

A die is thrown twice and the sum of numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared atleast once ? (NCERT)

Answer:

When a dice is thrown twice, total number of outcomes = 6 × 6 = 36

Let us consider the events:

A = number 4 appears atleast one = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}

B = Sum of the number be 6 = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)} A ∩ B = {(2, 4), (4, 2)}

A ∩ B = {(2, 4), (4, 2)}

Thus P (A ∩ B) = \(\frac{2}{36}\) ; P (B) = \(\frac{5}{36}\)

required probability = P (A/B)

= \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2 / 36}{5 / 36}=\frac{2}{5}\)

Question 6 (odd).

If P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{3}\) and P(A ∩ B) = \(\frac{1}{5}\), find P(Ā/B̄).

Answer:

Given P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{3}\) and P(A ∩ B) = \(\frac{1}{5}\)

Question 7.

A pair of dice is thrown and the sum of the numbers is observed to be even. What is the probability that both dice have come up with even numbers ?

Answer:

Let A : event that sum of the numbers to be even

B : event that both dice have come up with even numbers.

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)}

B = {(2, 2), (4, 4), (6, 6), (2, 4), (2, 6), (4, 2), (4, 6), (6, 2), (6, 4)}

∴ A ∩ B = {(2, 2), (2, 4), (2, 6), (4, 2), (4,6), (4,4),(6,6), (6,2), (6,4)}

Total no. of outcomes = 36

Thus required probability = P (B | A)

= \(\frac{P(A \cap B)}{P(A)}\)

Here P(A ∩ B) = \(\frac{9}{36}=\frac{1}{4}\)

and P(A) = \(\frac{18}{36}=\frac{1}{2}\)

∴ required probability = \(\frac{1 / 4}{1 / 2}\) = 1/2

Question 8.

Given that the numbers appearing on rolling two dice together are different. What is the probability that the sum of numbers appearing on two dice is 6 ?

Answer:

When two dice are thrown .-. n (S) = 62 = 36

E: event that sum of appearing on two dice is 6

F : event that numbers appearing on rolling two dice together are different

∴ E = {(1, 51, (2, 4), (3^ 3), (4, 2), (5, 1)}

F = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6, 5)}

∴ E ∩ F = {(1, 5), (2, 4), (4, 2), (5, 1)}

n (E ∩ F) = 4 : n (F) = 30

Thus required prob. = P (E/F) = \(\frac{P(E \cap F)}{P(F)}\)

= \(\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~F})}=\frac{4}{30}\)

= \(\frac{2}{15}\)

Question 9.

A family has two children.

(i) What is the probability that both the children are boys given that atleast one of them is a boy ?

(ii) What is the probability that both children are boys if it is known that the elder child is a boy ?

(iii) What is the probability that both children are girls if it is known that elder child is a girl ? (NCERT)

Answer:

(i) Thus sample space S = {BG, GB, BB, GG}

where B = Boy and G = Girl,

first letter = elder child and 2nd letter = younger child

Let E : both are boys ; F : atleast one of children is boy.

∴ E = {BB} ; F = {BQ GB, BB}

∴ E ∩ F = {BB}

P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{1 / 4}{3 / 4}=\frac{1}{3}\)

Question 10.

12 cards numbered 1 to 12 are placed in a box, mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, then find the probability that it is an even number.

Answer:

Let the events be :

E : number on card drawn is even

F : number on card drawn is > 3.

∴ E = {2, 4, 6, 8, 10, 12}

and F = {4, 5, 6, 7, 8, 9, 10, 11, 12}

E ∩ F = {4, 6, 8, 10, 12}

Thus required probability = P (E | F)

= \(\frac{P(E \cap F)}{P(F)}=\frac{5 / 12}{9 / 12}=\frac{5}{9}\)

Question 11.

Two digits are selected at random from the digits 1 to 9. If the sum is even, then find the probability that both digits are odd.

Answer:

Consider the following events :

A = both numbers are odd

B = getting the sum as even number

Here S = {1, 2, 3, 4, 5, 6, 7, 8, 9}

There are 5 odd integers and 4 even integers

∴ P(A) = \(\frac{{ }^5 \mathrm{C}_2}{{ }^9 \mathrm{C}_2}\)

Since the sum of two integers is even if either both are even or both are odd

∴ P(B) = \(\frac{{ }^5 \mathrm{C}_2+{ }^4 \mathrm{C}_2}{{ }^9 \mathrm{C}_2}\) and P(A ∩ B) = \(\frac{{ }^5 C_2}{{ }^9 C_2}\)

Thus, required probability

Question 12.

One card is drawn from a well-shuffled pack of 52 cards. If E is the event ‘the card drawn is a king or a queen’ and F is the event ‘the card drawn is a queen or an ace’, then find the probability of conditional event E/F.

Answer:

Given E : event that the card drawn is king or queen

F : event that the card drawn is queen or an ace

∴ P(F) = \(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}\)

[since there are 4 queens and 4 aces]

P (E t F) P (drawing a queen) = \(\frac{4}{52}\)

∴ required probability = P (E|F)

= \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{4}{52}}{\frac{8}{52}}=\frac{1}{2}\)

Question 13.

(i) I roll two dice and get a sum more than 9. What is the probability that the number on the first die is even ?

(ii) I roll two dice and get an even number on the first die. What is the probability that sum is more than 9 ?

Answer:

(i) Let A: event of getting a sum more than 9. B : event that the number on first die is even

∴ A = {(4,6), (5, 5), (6,4), (5,6), (6, 5), (6,6)}

B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Total no. of outcomes = 6^{2} = 36 = n (S)

A ∩ B = {(4, 6), (6, 4), (6, 5), (6, 6)}

∴ n (A) = 6 ; n (B) = 18 ; n (A ∩ B) = 4

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{36}\)

P(A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{4}{36}\)

Thus required probability

= P(B|A) = \(\frac{\frac{4}{36}}{\frac{6}{36}}=\frac{2}{3}\)

= \(\frac{\frac{4}{36}}{\frac{6}{36}}=\frac{2}{3}\)

(ii) Also, P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{18}{36}\)

required probability = P (A/B)

= \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{36}}{\frac{18}{36}}=\frac{2}{9}\)

Question 14.

In a certain school, 20% of the students failed in English, 15% of the students failed in Mathematics and 10% of the students failed in both English and Mathematics. A student is selected at random. If he passed in English, what is the probability that he also passed in Mathematics ?

Answer:

Let E: event that a student is passed in English M : event that a student is passed in Mathematics

Given P(Ē) = 20% = \(\frac{1}{5}\)

∴ P(E) = 1 – \(\frac{20}{100}=\frac{4}{5}\)

Question 15.

60% students read Hindi newspaper, 40% students read Tamil newspaper and 20% students read both Hindi and Tamil newspaper. Find the probability that a student selected at random reads

(i) Tamil newspaper given that he has already read Hindi newspaper.

(ii) Hindi newspaper given that he has already read Tamil newspaper.

(iii) Neither Hindi nor Tamil newspaper. What values are being promoted in this question ? (Value Based)

Answer:

Let the events be

A : students reads Hindi newspaper

B : students reads Tamil newspaper

∴ P(A) = \(\frac{60}{100}=\frac{3}{5}\); P(B) = \(\frac{40}{100}=\frac{2}{5}\)

andP(A ∩ B) = \(\frac{20}{100}=\frac{1}{5}\)

(i) required probability = P (B/A)

= \(\frac{P(B \cap A)}{P(A)}=\frac{1 / 5}{3 / 5}=\frac{1}{3}\)

(ii) required probability = P (A/B)

= \(\frac{P(A \cap B)}{P(B)}=\frac{1 / 5}{2 / 5}=\frac{1}{2}\)

(iii) P (neither Hindi nor Tamil)

= P(A^{c} ∩ B^{c}) = [P (A ∪ B)^{c}]

= 1 – P(A ∪ B)

= 1 – [P(A) + P(B) – P(A ∩ B)]

= 1 – \(\left[\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right]\)

= 1 – \(\frac{4}{5}=\frac{1}{5}\)

Question 16.

A committee of 4 students is selected from a group consisting of 7 boys and 4 girls. Find the probability that there are exactly 2 boys in the committee, given that atleast one girl must be there in the committee.

Answer:

Total no. of students = 7 + 4 = 11

∴ Total no. of selecting 4 students out of 11 = ^{11}C_{4}

Let A : event that atleast one girl is chosen

B : even that exactly 2 boys in the committee.

To find P (B/A)

Let A’ : event that no girl is chosen

∴ P (A’) = P (selecting all 4 boys) = \(\frac{{ }^7 C_4}{{ }^{11} C_4}\)

= \(\frac{7 \times 6 \times 5 \times 4}{11 \times 10 \times 9 \times 8}=\frac{7}{66}\)

∴ P(A) = 1 – P(A’) = 1 – \(\frac{7}{66}=\frac{59}{66}\)

Since B ⊂ A ⇒ A ∩ B = B

⇒ P (A ∩ B) = P (B)

= P (exactly 2 boys are chosen)

= P (selecting 2 boys and 2 girls)