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

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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