Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 22.

Answers (1)

Answer: P(E \mid F)=\frac{P(E) F J}{P(F)}=\frac{2}{6}=\frac{1}{3}

Hint: P(B \mid 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 \mid 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} &\mathrm{E}=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\} \\ &\mathrm{F}=\left\{\begin{array}{l} (1,5),(2,5),(3,5),(4,5),(6,5) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \end{array}\right\} \\ &\mathrm{E} \cap \mathrm{F}=\{(5,5),(5,6),(6,5)\} \\ &P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{3}{11} \end{aligned}

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