Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 13 maths textbook solution

Answers (1)

Answer:

 \frac{5}{34}

Hint:

 Use Baye’s theorem.

Given:

  The first operator A produce 10% defective items whereas the two other operate B, C 5% and 7% defective respectively. A is an on 50% of time, B on job 30% of time and C on job 20% of time. The defective item is produced by A.

Solution:

Let E1,E2,E3 respective event of time consumed by machine A, B and c for job.

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

Let X be event produce defective item

\begin{aligned} &P\left ( \frac{X}{E_1} \right )=1%=\frac{1}{100}\\ &P\left ( \frac{X}{E_2} \right )=5%=\frac{5}{100}\\ &P\left ( \frac{X}{E_3} \right )=7%=\frac{7}{100}\\ \end{aligned}

The probability that defective item was produce by A is given as

\begin{aligned} &P\left ( \frac{E_1}{A} \right ) \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}{100} }}{\frac{1}{2}\times \frac{1}{100}+\frac{3}{10}\times \frac{5}{100}+\frac{1}{5}\times \frac{7}{100}}\\ &=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{100}\left ( \frac{1}{2}+\frac{3}{2}+\frac{5}{7} \right )}\\ &=\frac{\frac{1}{2}}{\frac{17}{5}}\\ &=\frac{5}{34} \end{aligned}

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