Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 27

Answers (1)

Answer:

 \frac{4}{9}

Hint:

 Use Baye’s theorem.

Given:

 There are three coins one is two headed coin, another is blazed coin that comes up heads 75% of time and third is unbiased coin.

Solution:

Let

C1= two headed coin

C2= biased coin

C3= unbiased coin

We need to find probability that coin is two headed it show

P\left (\frac{C_1}{H} \right )

\begin{aligned} &P(C_1)=P(C_2)=P(C_3)=\frac{1}{3}\\ &P\left ( \frac{A}{C_1} \right )=1\\ &P\left ( \frac{A}{C_2} \right )=\frac{3}{4}\\ &P\left ( \frac{A}{C_3} \right )=\frac{1}{2}\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{C_1}{H} \right )=\frac{P(C_1).P\left ( \frac{H}{C_1} \right )}{P(C_1)\times P\left ( \frac{H}{C_1} \right )+P(C_2)\times P\left ( \frac{H}{C_2} \right )+P(C_3)\times P\left ( \frac{H}{C_3} \right )}\\ &=\frac{{\frac{1}{3}\times 1 }}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}\\ &=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\left ( 1+\frac{3}{4}+\frac{1}{2} \right )}\\ &=\frac{1}{\frac{9}{4}}\\ &=\frac{4}{9} \end{aligned}

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