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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 3

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 3

Answers (1)

Answer:

 \frac{2}{9}

Hint:

 Use baye’s theorem.

Given:

 Three win contain 2 white and 3 black balls; 3 white, 2 black and 4 white and 1 black balls. One ball drawn from a win chosen at random and it was found to white. Find the probability of win drawn.

Solution:

Let E1,E2,E3  denote the event of selecting win I, win II, win III.

Let A be event that ball drawn white.

\begin{aligned} &P(E_1)=\frac{1}{3},\\ &P(E_2)=\frac{1}{3},\\ &P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\frac{2}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{3}{5}\\ &P\left ( \frac{A}{E_3} \right )=\frac{4}{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 )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times \frac{2}{5}}}{\frac{1}{3}\times \frac{2}{5}+\frac{1}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{4}{5}}\\ &=\frac{\frac{2}{5}}{\frac{2+3+4}{5}}\\ &=\frac{2}{9} \end{aligned}

Posted by

Gurleen Kaur

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