RD Sharma Class 12 Exercise 30.3 Probability Solutions Maths-Download PDF Online

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RD Sharma books are widely considered as the best material for maths as they are comprehensive and provide a vast number of concepts.. The Class 12 RD Sharma chapter 30 exercise 30.3 arrangement is one of the examples for it.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.3
RD Sharma Chapter-wise Solutions

Probability Excercise: 30.3

Probability Exercise 30.3 Question 1

Answer: $\frac{4}{9}$
Hint: Just simplify and use formula of probability
Given:

\begin{aligned} P (A) = \frac{7}{13} \\ P (B) = \frac{9}{13} \\ P (A \cap B) = \frac{4}{13} \end{aligned}

Solution:
Using,

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

Required Probability,

P (A ∣ B) = \frac{\frac{4}{18}}{\frac{9}{18}} = \frac{4}{9}

Probability Exercise 30.3 Question 2

Answer: $\frac{2}{3}, \frac{1}{3}$
Hint: $Probability = \frac{No. of outcomes}{Total number of outcomes}$
Given:

\begin{aligned} P (A) = 0.6 \\ P (B) = 0.3 \\ P (A \cap B) = 0.2 \end{aligned}

Solution:
Now,

\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} \\ = \frac{0.2}{0.3} \\ = \frac{2}{3} \\ P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \\ = \frac{0.2}{0.6} \\ = \frac{1}{3} \end{aligned}

Probability Exercise 30.3 Question 3

Answer: 0.64
Hint: $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$
Given:
$\begin{aligned} P (B) = 0.5 \\ P (A \cap B) = 0.32 \end{aligned}$
Solution:

\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} \\ = \frac{0.32}{0.5} \\ = 0.64 \end{aligned}

Probability Exercise 30.3 Question 4

Answer: 0.96
Hint: Use

P (A \cup B) = P (A) + P (B) - P (A \cap B)

Given,
$\begin{aligned} P (A) = 0.4 \\ P (B) = 0.8 \\ P (\frac{B}{A}) = 0.6 \end{aligned}$
Solution:
Now,

\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} \\ P (A \cap B) = 0.24 \\ \Rightarrow P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} \\ = \frac{0.24}{0.8} \\ = 0.3 \\ \Rightarrow P (A \cup B) = P (A) + P (B) - P (A \cap B) 4 \\ = 0.4 + 0.8 - 0.24 \\ = 0.96 \end{aligned}

Probability Exercise 30.3 Question 5(i)

Answer: $\frac{2}{3}, \frac{1}{2}$
Hint:
$\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} \\ P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \end{aligned}$
Given,
$\begin{aligned} P (A) = \frac{1}{3} \\ P (B) = \frac{1}{4} \\ P (A \cup B) = \frac{5}{12} \end{aligned}$
$\begin{aligned} P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ P (A \cap B) = P (A) + P (B) - P (A \cup B) \end{aligned}$
$\begin{aligned} P (A \cap B) = \frac{1}{3} + \frac{1}{4} - \frac{5}{12} \\ P (A \cap B) = \frac{2}{12} \\ P (A \cap B) = \frac{1}{6} \end{aligned}$
Solution:
Now,

\begin{matrix} P (A ∣ B) = \frac{P (A \cap B)}{p (B)} \\ = \frac{\frac{1}{6}}{\frac{1}{4}} \\ = \frac{4}{6} = \frac{2}{3} \end{matrix}

\begin{matrix} P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \\ = \frac{\frac{1}{6}}{\frac{1}{3}} \\ = \frac{1}{2} \end{matrix}

Probability Exercise 30.3 Question 5(ii)

Answer:

\frac{4}{5}, \frac{2}{3}

Hint:
$\begin{aligned} P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ P (A \cap B) = P (A) + P (B) - P (A \cup B) \end{aligned}$
Solution:
Given,
$\begin{aligned} P (A) = \frac{6}{11} \\ P (B) = \frac{5}{11} \\ P (A \cup B) = \frac{7}{11} \end{aligned}$
Now,

\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} \\ = \frac{\frac{4}{11}}{\frac{5}{11}} \\ = \frac{4}{5} \end{aligned}

\begin{aligned} P (B ∣ A) & = \frac{P (A \cap B)}{P (A)} \\ = \frac{\frac{4}{11}}{\frac{6}{11}} \end{aligned}

= \frac{4}{6}

= \frac{2}{3}

Probability Exercise 30.3 Question 5(iii)

Answer: $\frac{5}{9}$

Hint: $P (\bar{A} \cap B) = P (B) - P (A \cap B)$
Given:
$\begin{aligned} P (A) = \frac{7}{13} \\ P (B) = \frac{9}{13} \\ P (A \cap B) = \frac{4}{13} \end{aligned}$
Solution:

\begin{aligned} P (\bar{A} \cap B) = P (B) - P (A \cap B) \\ = \frac{9}{13} - \frac{4}{13} \\ = \frac{5}{13} \end{aligned}

Now,

\begin{aligned} P (\bar{A} / B) = \frac{P (\bar{A} \cap B)}{P (B)} \\ = \frac{\frac{5}{13}}{\frac{9}{13}} \\ = \frac{5}{9} \end{aligned}

Probability Exercise 30.3 Question 5(iv)

Answer: $\frac{1}{4}, \frac{5}{8}$
Hint: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Given
$\begin{aligned} P (A) = \frac{1}{2} \\ P (B) = \frac{1}{3} \end{aligned}$
Solution:
$P (A \cap B) = \frac{1}{4}$
Also,

\begin{aligned} P (\overset{―}{B}) = 1 - P (B) \\ = 1 - \frac{1}{3} \\ = \frac{2}{3} \\ \Rightarrow P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\ = \frac{6 + 4 - 3}{12} \\ = \frac{7}{12} \end{aligned}

Also,

\begin{aligned} P (\bar{A} \cap B) = P (B) - P (A \cap B) \\ = \frac{1}{3} - \frac{1}{4} \\ = \frac{4 - 3}{12} \\ P (\bar{A} \cap B) = \frac{1}{12} \end{aligned}

Solution:
We know

\begin{aligned} P (\bar{A} \cap \bar{B}) = P (\overset{―}{A \cup B}) \\ = 1 - P (A \cup B) \\ = 1 - \frac{7}{12} \\ = \frac{5}{12} \end{aligned}

\begin{aligned} P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} \\ = \frac{\frac{1}{4}}{\frac{1}{3}} \\ = \frac{3}{4} \end{aligned}

\begin{aligned} P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \\ = \frac{\frac{1}{4}}{\frac{1}{2}} \\ = \frac{2}{4} = \frac{1}{2} \end{aligned}

\begin{aligned} P (\frac{\bar{A}}{B}) = \frac{P (\bar{A} \cap B)}{P (B)} \\ = \frac{\frac{1}{12}}{\frac{1}{3}} \\ = \frac{3}{12} = \frac{1}{4} \\ P (\frac{\bar{A}}{\bar{B}}) = \frac{P (\bar{A} \cap \bar{B})}{P (\bar{B})} \end{aligned}

\begin{aligned} = \frac{\frac{5}{12}}{\frac{2}{3}} \\ = \frac{15}{24} = \frac{5}{8} \end{aligned}

Probability Exercise 30.3 Question 6

Answer: $\frac{11}{26}$
Hint: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Given,
$\begin{aligned} 2 P (A) = P (B) = \frac{5}{13} \\ P (A ∣ B) = \frac{2}{5} \\ P (A) = \frac{5}{26} \\ P (B) = \frac{5}{13} \end{aligned}$
Solution:
Now,

\begin{aligned} P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} \\ \frac{2}{5} = \frac{P (A \cap B)}{\frac{5}{13}} \\ P (A \cap B) = \frac{2}{5} \times \frac{5}{13} \end{aligned}

= \frac{2}{13}

\Rightarrow P (A \cup B) = P (A) + P (B) - P (A \cap B)

\begin{aligned} = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \\ = \frac{11}{26} \end{aligned}

Probability Exercise 30.3 Question 7

Answer:

\frac{4}{11}, \frac{4}{5}, \frac{2}{3}

Hint: Use

\Rightarrow P (A \cup B) = P (A) + P (B) - P (A \cap B)

Given,
$\begin{aligned} P (A) = \frac{6}{11} \\ P (B) = \frac{5}{11} \\ P (A \cup B) = \frac{7}{11} \end{aligned}$
Solution:
Now, (i)

\begin{aligned} P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P (A \cap B) \\ P (A \cap B) = 1 - \frac{7}{11} \\ P (A \cap B) = \frac{4}{11} \end{aligned}

(ii)

\begin{aligned} P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} \\ = \frac{\frac{4}{11}}{\frac{5}{11}} \\ = \frac{4}{5} \end{aligned}

(iii)

\begin{aligned} P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \\ = \frac{\frac{4}{11}}{\frac{6}{11}} \\ = \frac{4}{6} = \frac{2}{3} \end{aligned}

Probability Exercise 30.3 Question 8(i)

Answer: $\frac{1}{2}$
Hint: $Probability = \frac{No. of Outcomes}{Total no of outcomes}$
Given, A=Heads on third toss
B=Heads on first two tosses
Solution:
$\begin{aligned} S = {(H H H), (H H T), (H H T), (T T H), (T T T), (H T H), (T H H), (T T H)} \\ A = {(H H H), (H T H), (T H H), (T T H)} \\ B = {(H H H), (H H T)} \\ A \cap B = {H H H} \end{aligned}$
$P (A) = 4, P (B) = 2, P (A \cap B) = 1$
Required Probability

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

\begin{aligned} = \frac{P (A \cap B)}{P (B)} \\ = \frac{1}{2} \end{aligned}

Probability Exercise 30.3 Question 8(ii)

Answer: $= P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{3}{7}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given, A=At least two heads
B=At most two heads
Solution:
Clearly,

S = {(H H H), (H H T), (H H T), (T T H), (T T T), (H T H), (T H H), (T T H)}

\begin{aligned} A = {(HHH), (HTH), (THH), (HHT)} \\ B = {(TT), (HTH), (HTT), (THT), (THH), (HHT), (HTT)} \\ A \cap B = {(HTH), (THH), (HHT)} \end{aligned}

P (A) = 4

P (B) = 7

Required Probability $= P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{3}{7}$

Probability Exercise 30.3 Question 8(iii)

Answer : $= P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{6}{7}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given,
A=At most two tails
B=At least one tail
Solution:
Clearly,

\begin{aligned} S = {(H H H), (H H T), (H H T), (T T H), (T T T), (H T H), (T H H), (T T H)} \\ A = {(π H), (T H H), (HHT), (T H T), (HHT), (HTT), (HHH)} \\ B = {(TT), (TH), (THH), (HHT), (THT), (HHT), (HTT)} \end{aligned}

Now,

A \cap B = {(TTH), (THH), (HHT), (HTT)}

P(A) = 7
P(B) = 7
Required Probability $= P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{6}{7}$

Probability Exercise 30.3 Question 9(i)

Answer: 1
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given, A=Tail appears on one coin
B=One coin shows head
Solution:
Clearly,

\begin{aligned} A = {(H, T), (T, H)} \\ B = {(H, T), (T, H)} \end{aligned}

P(A) = 2
P(B) = 2
Now,

A \cap B = {(H, T), (T, H)}

Required Probability

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

Probability Exercise 30.3 Question 9(ii)

Answer: 0
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given events,
A=No tail appears
B=No head appears
Solution:
Clearly,

\begin{aligned} A = {(H, H)} \\ B = {(T, T)} \end{aligned}

\begin{aligned} A \cap B = {ϕ} \\ P (A) = 2 \\ P (B) = 2 \end{aligned}

Required Probability

= P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = 0

Probability Exercise 30.3 Question 10

Answer: $\frac{1}{6}, \frac{1}{36}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given,
A=Getting 4 on trial throw
B=Getting 6 on first throw and 5 on second throw
Solution:
Clearly,

\begin{matrix} A = {\begin{cases} (1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), \\ (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), \\ (2, 5, 4), (2, 6, 4), (3, 1, 4), (3, 2, 4), (3, 3, 4), \\ (3, 4, 4), (3, 5, 4), (3, 6, 4), (4, 1, 4), (4, 2, 4), \\ (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4), (5, 1, 4), \\ (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4), \\ (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), \\ (6, 6, 4) \end{cases}} \\ B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)} \end{matrix}

Now,

A \cap B = {(6, 5, 4)}

P(A) = 36
P(B) = 6

\begin{aligned} P (A ∣ B) = \frac{P (A \cap B)}{P (B)} = \frac{1}{6} \\ P (B ∣ A) = \frac{P (A \cap B)}{P (A)} = \frac{1}{36} \end{aligned}

Probability Exercise 30.3 Question 11

Answer: $1, \frac{1}{2}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given, A= Son standing on one end
B= Father standing in the middle
Solution:
Clearly,
S=Total events

\begin{aligned} = {MFS, MSF, FSM, FMS, SMF, SFM} \\ A = {MFS, FMS, SMF, SFM} \\ B = {MFS, SFM} \\ A \cap B = {MFS, SFM} \end{aligned}

(1)

\begin{matrix} P (\frac{A}{B}) = \frac{n (A \cap B)}{n (B)} \\ = \frac{2}{2} \\ = 1 \end{matrix}

(2)

\begin{matrix} P (B ∣ A) = \frac{P (A \cap B)}{P (A)} \\ = \frac{2}{4} \\ = \frac{1}{2} \end{matrix}

Probability Exercise 30.3 Question 12

Answer: $\frac{2}{5}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given,
A= 4 appears on the die at least once
B=The sum of the numbers on two dice is 6
Solution:

\begin{aligned} A = {\begin{matrix} (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), \\ (4, 4), (4, 5), (4, 6), (5, 4), (6, 4) \end{matrix}} \\ B = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)} \\ A \cap B = {(2, 4), (4, 2)} \end{aligned}

P(A) = 11
P(B) = 5
Required Probability

\begin{aligned} = P (\frac{A}{B}) \\ = \frac{n (A \cap B)}{n (B)} = \frac{2}{5} \end{aligned}

Probability Exercise 30.3 Question 13

Answer: $\frac{1}{6}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$

Solution:
Clearly,

\begin{aligned} A = {\begin{matrix} (1, 4), (2, 4), (3, 4), \\ (4, 4), (5, 4), (6, 4) \end{matrix}} \\ B = {(4, 4), (3, 5), (5, 3), (2, 6), (6, 2)} \\ A \cap B = {(4, 4)} \end{aligned}

Required Probability =

P (B ∣ A) = \frac{P (A \cap B)}{P (A)} = \frac{1}{6}

Probability Exercise 30.3 Question 14

Answer: $= P (B ∣ A) = \frac{P (A \cap B)}{p (A)} = \frac{1}{6}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given,
A=No. appearing on second die is odd.
B=The sum of the no. on two dice is 7.
Solution:
Clearly,

\begin{aligned} A = {\begin{matrix} (1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), \\ (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5) \\ (5, 1), (5, 3), (5, 5), (6, 1), (6, 3), (6, 5) \end{matrix}} \\ B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)} \\ A \cap B = {(2, 5), (4, 3), (6, 1)} \end{aligned}

Required Probability

= P (B ∣ A) = \frac{P (A \cap B)}{p (A)} = \frac{1}{6}

Probability Exercise 30.3 Question 15

Answer: $P (B ∣ A) = \frac{P (A \cap B)}{p (A)} = \frac{1}{6}$

Hint: Use,

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

Given,
A= Prime number appears on second die
B=The sum of the numbers on two dice is 7
Solution:
Clearly,

\begin{aligned} A = {\begin{matrix} (1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), \\ (3, 2), (3, 3), (3, 5), (4, 2), (4, 3), (4, 5), \\ (5, 2), (5, 3), (5, 5), (6, 2), (6, 3), (6, 5) \end{matrix}} \\ B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)} \\ A \cap B = {(2, 5), (5, 2), (4, 3)} \end{aligned}

P(A) = 18
P(B) = 6
Required Probability

P (B ∣ A) = \frac{P (A \cap B)}{p (A)} = \frac{1}{6}

Probability Exercise 30.3 Question 16

Answer: $\frac{2}{3}$
Hint: Use,

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

Given,
A= No. is odd
B=No. is prime
Solution:
Clearly,

\begin{aligned} A = {1, 3, 5} \\ B = {2, 3, 5} \\ A \cap B = {3, 5} \end{aligned}

Required Probability

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

= \frac{2}{3}

Probability Exercise 30.3 Question 17

Answer: $P (B ∣ A) = \frac{n (A ∣ B)}{n (A)} = \frac{3}{6} = \frac{1}{2}$
Hint: Use,

P (B ∣ A) = \frac{P (A ∣ B)}{P (A)}

Given,
A = 4 appears on first die
B = The sum of the numbers is on two dice is 8 or more
Solution:

\begin{aligned} A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)} \\ n (A) = 6 \\ B = {\begin{matrix} (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), \\ (4, 6), (5, 3), (5, 4), (5, 6), (6, 2), \\ (6, 3), (6, 4), (6, 5), (6, 6) \end{matrix}} \\ n (B) = 15 \end{aligned}

Now,

\begin{aligned} P (A) = 6 \\ P (B) = 15 \\ A \cap B = {(4, 4), (4, 5), (4, 6)} \end{aligned}

Required Probability

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

Probability Exercise 30.3 Question 18

Answer: $\frac{3}{25}$
Hint: $P (B ∣ A) = \frac{P (A \cap B)}{P (A)}$
Given,
A=At least one die does not show 5
B=The sum of the numbers on two dice is 8
Solution:
Clearly,

$\begin{aligned} A = {\begin{matrix} (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), \\ (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), \\ (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), \\ (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), \\ (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) \end{matrix}} \\ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} \\ A \cap B = {(4, 4), (6, 2), (2, 6)} \end{aligned}$
Required Probability

= P (B ∣ A) = \frac{P (A \cap B)}{P (A)} = \frac{3}{25}

Probability Exercise 30.3 Question 19

Answer: $\frac{10}{5} = \frac{5}{8}$
Hint: Use combination method of

n_{C_{r}} = \frac{n!}{(n - r)! r!}

Given: O represents the event of getting two odd numbers
S represents the event of getting their sum as an even number.
Solution:

\begin{matrix} P (S) = (4 C_{2} + 5 C_{2}) / 9 C_{2} \\ P (O ∣ S) = \frac{n (O \cap S)}{n (S)} \\ = \frac{\frac{s_{2}}{9} C_{2}}{\frac{(^{4} C_{2} + 5_{2})}{^{9} C_{2}}} \end{matrix}

= \frac{5 c_{2}}{(4 c_{2} + 5_{c 2})} {(5_{C_{2}}) = \frac{5 \times 4 \times 3 \times 2 \times 1}{(5 - 2)! (2 \times 1)} = \frac{120}{3 \times 2 \times 1 \times 2} = 10}

$\frac{10}{5} = \frac{5}{8}$

Probability Exercise 30.3 Question 20

Answer: $\frac{2}{5}$
Hint: $P (A ∣ B) = \frac{n (A \cap B)}{n (B)}$
Given :- Now, Let A = 5 appears on the die at least once.
B = The sum of the no. on two dice is 8
Solution:
Clearly,

\begin{aligned} A = {\begin{matrix} (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), \\ (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), \\ (5, 6) \end{matrix}} \\ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} \\ A \cap B = {(3, 5), (5, 3)} \end{aligned}

P(A) = 11
P(B) = 5
Required Probability $P (A ∣ B) = \frac{P (A \cap B)}{P (B)} = \frac{2}{3}$

Probability Exercise 30.3 Question 21

Answer: $= \frac{1}{6}$
Hint:

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

Given :- Let, A=First die shows
B=The sum of the numbers on two dice is 7
Solution:
Clearly,

\begin{aligned} A = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} \\ B = {(2, 5), (5, 2), (4, 3), (3, 4), (1, 6), (6, 1)} \\ A \cap B = {(6, 1)} \end{aligned}

P(A) = 6
P(B) = 6
Required Probability

= P (B ∣ A) = \frac{P (A \cap B)}{p (A)} = \frac{1}{6}

Probability Exercise 30.3 Question 22

Answer: $P (E ∣ F) = \frac{P (E) F J}{P (F)} = \frac{2}{6} = \frac{1}{3}$
Hint: $P (B ∣ A) = \frac{P (A \cap B)}{P (A)}$
Given, E = The sum of the no. on two dice is 10 or more.
F = 5 appears on first die
Solution:
Clearly,

\begin{aligned} E = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} \\ F = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} \\ E \cap F = {(5, 5), (5, 6)} \end{aligned}

P(F) = 6
P(E) = 6

= P (E ∣ F) = \frac{p (E \cap F)}{p (F)} = \frac{2}{6} = \frac{1}{3}

Second case:
Let, E = The sum of the no. on two dice is 10 or more.
F = 5 appears on at least in one die
Now, P(F) = 11
P(E) = 6

\begin{aligned} E = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} \\ F = {\begin{matrix} (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) \\ (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) \end{matrix}} \\ E \cap F = {(5, 5), (5, 6), (6, 5)} \\ P (E ∣ F) = \frac{P (E \cap F)}{P (F)} = \frac{3}{11} \end{aligned}

Probability Exercise 30.3 Question 23

Answer: $\frac{5}{8}$
Hint: $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$
Given :- Let, M= Students passes Mathematics,
C=Students Passes Computer Science

\begin{aligned} P (M) = \frac{4}{5} \\ P (M \cap C) = \frac{1}{2} \end{aligned}

Now,

\begin{matrix} P (C ∣ M) = \frac{P (CnM]}{p (M)} \\ = \frac{\frac{1}{2}}{\frac{4}{5}} \\ = \frac{5}{8} \end{matrix}

Probability Exercise 30.3 Question 24

Answer: 0.6
Hint: $P (E ∣ F) = \frac{P (E \cap F)}{p (F)}$
Given, S=Person buying a shirt
T=Person buying a trouser
Solution:

\begin{aligned} P (S) = 0.2 \\ P (T) = 0.3 \\ P (S ∣ T) = 0.4 \\ P (S ∣ T) = \frac{P (S \cap T)}{P (T)} \\ P (S \cap T) = P (S ∣ T) \cdot P (T) \\ = 0.4 \times 0.3 \\ = 0.12 \end{aligned}

Now,

\begin{aligned} P (T ∣ S) = \frac{P (S \cap T)}{P (S)} \\ = \frac{0.12}{0.2} = 0.6 \end{aligned}

Probability Exercise 30.3 Question 25

Answer: $\frac{1}{10}$
Hint: $P (\frac{S}{G}) = \frac{P (S \cap G)}{P (G)}$
Given, S= Student chosen randomly studies in class XII
G= Female student chosen randomly

\begin{aligned} P (G) = \frac{430}{1000} \\ P (S \cap G) = \frac{43}{1000} \\ P (\frac{S}{G}) = \frac{P (S \cap G)}{P (G)} \end{aligned}

\begin{array}{r} = \frac{\frac{430}{1000}}{\frac{430}{1000}} \\ = \frac{1}{10} \end{array}

Probability Exercise 30.3 Question 26

Answer: $\frac{4}{7}$
Hint: $Probability = \frac{No. of outcomes}{Total no.of Outcomes}$
Given:
Total Possibility of cards = {1,2,3,4,5,6,7,8,9,10}
Solution:
Let A = An even number on the card
B = A number more than 3 on the card.

\begin{aligned} A = {2, 4, 6, 8, 10} \\ B = {4, 5, 6, 7, 8, 9, 10} \\ A \cap B = {4, 6, 8, 10} \end{aligned}

P(A) = {5}
P(B) = {7}

Required probability

P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)} = \frac{4}{7}

Probability Exercise 30.3 Question 27

Answer: $\frac{1}{2}, \frac{1}{3}$
Hint:

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

Given, A=Both the children are girls
B=The youngest child is a girl
C=At least one child is a girl
Solution:

\begin{aligned} S = {B_{1} B_{2}, B_{1} G_{2}, G_{1} B_{2}, G_{1} G_{2}} \\ A = {G_{1} G_{2}, G_{2} G_{1}} \\ B = {B_{1} G_{2}, G_{1} G_{2}} \\ C = {B_{1} G_{2}, G_{1} B_{2}, G_{1} G_{2}} \\ A \cap B = {G_{1} G_{2}} \\ A \cap C = {G_{1} G_{2}} \\ P (A) = 2 \\ P (B) = 2 \end{aligned}

Required probability $P (A ∣ B) = \frac{P (A \cap B)}{p (B)} = \frac{1}{2}$
Required probability $P (A ∣ C) = \frac{P (A \cap C)}{p (C)} = \frac{1}{3}$

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