Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 36 maths

Answers (1)

Answer:

 \frac{4}{9}

Hint:

 Use baye’s theorem.

Given:

 A speaks the truth 8 time out of 10 time a die is tossed. He report that it was 5.

Solution:

Let A denote the event that man reports 5 occur and E the event that actually 5 tossed up.

\begin{aligned} &P(E)=\frac{1}{6}\\ &P(E)=1-\frac{1}{6}=\frac{5}{6}\\ \end{aligned}

Also,

\begin{aligned} &P\left ( \frac{A}{E} \right ) \end{aligned}

=Probability that man reports that 5 occur given that 5 actually turned up

= probability of man speak the truth

\begin{aligned} &=\frac{8}{10}=\frac{4}{5} \end{aligned}

Probability that man reports that 5 occur given that 5 does not turned up

= probability of man not speak the truth

\begin{aligned} &=1-\frac{4}{5}=\frac{1}{5} \end{aligned}

Using Baye’s theorem we get

Required probability

\begin{aligned} &P\left (\frac{E}{A} \right )=\frac{P(E_).P\left ( \frac{A}{E} \right )}{P(E)\times P\left ( \frac{A}{E} \right )+P(E)\times P\left ( \frac{A}{E} \right )}\\ &=\frac{\frac{1}{6}\times \frac{4}{5}}{\frac{1}{6}\times \frac{4}{5}+\frac{5}{6}\times \frac{1}{5}} \\ &=\frac{4}{9} \end{aligned}

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