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  • Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 1 maths textbook solution

Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 1 maths textbook solution

Answers (1)

Answer:

 33/118, 55/118, 30/118

Hint:

 Use baye’s theorem.

Given:

The content of win I, II, III is as follows

Win I; 1 white, 2 black and 3 red balls.

Win II: 1 black, 2 white and 1 red balls,

Win III: 4 white, 5 black, 3 red balls

One win is chosen at random and two balls are drawn. The happen to be white and red what is probabilities that come from win I, II, III.

Solution:

Let E1,E2,E3  be selecting win I, win II, and win III.

Let A be event of chosen two ball (w,R).

\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{1C_1\times 3C_1}{6C_2}\\ &=\frac{1\times 3}{15}\\ &=\frac{1}{5}\\ &P\left (\frac{A}{E_2} \right )=\frac{2C_1\times 1C_1}{4C_2}\\ &=\frac{2\times 1}{6}\\ &=\frac{1}{3}\\ \end{aligned}

\begin{aligned} &P\left (\frac{A}{E_3} \right )=\frac{4C_1\times 3C_1}{12C_2}\\ &=\frac{4\times 3}{66}\\ &=\frac{2}{11}\\ \end{aligned}

Using Bayes’ 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).P\left ( \frac{A}{E_1} \right )+P(E_2).P\left ( \frac{A}{E_2} \right )+P(E_3).P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}.\left ( \frac{1}{5} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{1}{5}}{\frac{37+55+30}{165}}\\ &=\frac{1}{5}\times \frac{165}{118}\\ &=\frac{33}{118} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \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}.\left ( \frac{1}{3} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{1}{3}}{\frac{37+55+30}{165}}\\ &=\frac{1}{3}\times \frac{165}{118}\\ &=\frac{55}{118} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \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}.\left ( \frac{2}{11} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{2}{11}}{\frac{37+55+30}{165}}\\ &=\frac{2}{11}\times \frac{165}{118}\\ &=\frac{30}{118} \end{aligned}

Posted by

Gurleen Kaur

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