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

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}

Posted by

Gurleen Kaur

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