Question

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?

Answer 1

Let E₁, E₂ and E₃ be the time taken by machine operators A, B, and C, respectively.

Let X be the event of producing defective items.

$$\therefore P\left(E_1\right)=50$$

$$P\left(E_2\right)=30$$

$$P\left(E_3\right)=20$$

$$\text{ Now },$$

$$P\left(A/E_1\right)=1$$

$$P\left(A/E_2\right)=5$$

$$P\left(A/E_2\right)=7$$

$$\text{ Using Bayes' theorem, we get }$$

$$\text{ Required probability }=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)++P\left(E_2\right)P\left(A/E_2\right)}$$

$$=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{2}\times \frac{1}{100}+\frac{3}{10}\times \frac{5}{100}+\frac{1}{5}\times \frac{7}{100}}$$

$$=\frac{5}{34}$$