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  • Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 4 maths

Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 4 maths

Answers (1)

Answer:

 \begin{aligned} &\frac{1}{40} \end{aligned}

Hint:

 Use baye’s theorem.

Given:

The content of three win

Win I: 7 white, 3 black;

Win II: 4 white, 6 black

Win III: 2 white 8 black

One ball drawn as random with probability 0.20, 0.60, 0.20 respectively drawn chosen win two balls drawn random without replacement. If both these ball drawn what is probability came from win III?

Solution:

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

Let A be event of two balls drawn white.

\begin{aligned} &P(E_1)=\frac{20}{100}\\ &P(E_2)=\frac{60}{100}\\ &P(E_3)=\frac{20}{100}\\ &P\left ( \frac{A}{E_1} \right )=\frac{7C_2}{10C_2}=\frac{21}{45}\\ &P\left ( \frac{A}{E_2} \right )=\frac{4C_2}{10C_2}=\frac{6}{45}\\ &P\left ( \frac{A}{E_3} \right )=\frac{2C_2}{10C_2}=\frac{1}{45}\\ \end{aligned}

Using Baye’s theorem we get

\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{20}{100}\times \frac{1}{45} }}{\frac{20}{100}\times \frac{21}{45}+\frac{60}{100}\times \frac{6}{45}+\frac{20}{100}\times \frac{1}{45}}\\ &=\frac{1}{21+18+1}\\ &=\frac{1}{40} \end{aligned}

 

 

 

 

 

 

Posted by

Gurleen Kaur

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