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

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}

Posted by

Gurleen Kaur

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