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  • Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 18

Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 18

Answers (1)

Answer:

 \frac{36}{61}

Hint:

 Use Baye’s theorem.

Given:

 Three win A, B, C contain 6 red and 4 white; 2 red, 6 white and 1 red 5 white ball respectively. A win is chosen at random and ball is drawn. If ball drawn is found to red. Find the probability that ball from win A.

Solution:

Let E1 and  E2  denotes the event that ball is red, bag A chose B chosen and bag C chosen respectively.

\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{6}{10}=\frac{3}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{2}{8}=\frac{1}{4}\\ &P\left ( \frac{A}{E_3} \right )=\frac{1}{6}\\ \end{aligned}

Using Baye’s theorem we get

\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{3}{5} }}{\frac{1}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{1}{4}+\frac{1}{3}\times \frac{1}{6}}\\ &=\frac{\frac{3}{5}}{\frac{3}{5}+\frac{1}{4}+\frac{1}{6}}\\ &=\frac{36}{61} \end{aligned}

Posted by

Gurleen Kaur

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