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}

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