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Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 17.

Answers (1)

I infoexpert24

Answer: $P(B \mid A)=\frac{n(A \mid B)}{n(A)}=\frac{3}{6}=\frac{1}{2}$

Hint: Use, $P(B \mid A)=\frac{P(A \mid B)}{P(A)}$

Given,

A = 4 appears on first die

B = The sum of the numbers is on two dice is 8 or more

Solution:

$\begin{aligned} &A=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\} \\ &n(A)=6 \\ &B=\left\{\begin{array}{l} (2,6),(3,5),(3,6),(4,4),(4,5), \\ (4,6),(5,3),(5,4),(5,6),(6,2), \\ (6,3),(6,4),(6,5),(6,6) \end{array}\right\} \\ &n(B)=15 \end{aligned}$

Now,

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

Required Probability $P(B \mid A)=\frac{P(A \mid B)}{P(A)}=\frac{1}{2}$

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