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  • Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 14

Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 14

Answers (1)

Answer:

 \frac{5}{11}

Hint:

 Use Baye’s theorem.

Given:

  50% of manufacture on machine A, 30% on B and 20% on C. 2% of item produced on A and 2% item produce on B are defective and 3% of these produce on C.

Solution:

Let

E1= event that item is manufacture on A

E2= event that item is manufacture on B

E3= event that item is manufacture on C

Let E be event that item defective.

\begin{aligned} &P(E_1)=\frac{50}{100}=\frac{1}{2}\\ &P(E_2)=\frac{30}{100}=\frac{3}{10}\\ &P(E_3)=\frac{20}{100}=\frac{1}{5}\\ \end{aligned}

\begin{aligned} &P\left ( \frac{X}{E_1} \right )=\frac{2}{100}=\frac{1}{50}\\ &P\left ( \frac{X}{E_2} \right )=\frac{2}{100}=\frac{1}{50}\\ &P\left ( \frac{X}{E_3} \right )=\frac{3}{100}\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{X} \right )=\frac{P(E_1).P\left ( \frac{X}{E_1} \right )}{P(E_1)\times P\left ( \frac{X}{E_1} \right )+P(E_2)\times P\left ( \frac{X}{E_2} \right )+P(E_3)\times P\left ( \frac{X}{E_3} \right )}\\ &=\frac{{\frac{1}{2}\times \frac{1}{50} }}{\frac{1}{2}\times \frac{1}{50}+\frac{3}{10}\times \frac{1}{50}+\frac{1}{5}\times \frac{3}{100}}\\ &=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}\\ &=\frac{\frac{1}{100}}{\frac{5+3+3}{500}}\\ &=\frac{5}{11} \end{aligned}

Posted by

Gurleen Kaur

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