Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 35

Answers (1)

Answer:

 \frac{3}{13}

Hint:

 Use baye’s theorem.

Given:

 A is known to speak with 3 time out of 5 time. He throws a die and report that is one. Find the probability that actually one.

Solution:

Let  A, E1 and E2  denote the events that the man reports the appearance of 1 on throwing a die, 1 occur and 1 does not occur respectively.

\begin{aligned} &P(E_1)=\frac{1}{6}\\ &P(E_2)=\frac{5}{6}\\ \end{aligned}

Now,

\begin{aligned} &P\left ( \frac{A}{E_1} \right )=\frac{3}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{2}{5}\\ \end{aligned}

Using Baye’s theorem we get

Required probability

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{1}{6}\times \frac{3}{5} }}{\frac{1}{6}\times \frac{3}{5}+\frac{5}{6}\times \frac{2}{5}}\\ &=\frac{3}{3+10}\\ &=\frac{3}{13} \end{aligned}

 

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