A manufacturer has three machine operators $A, B$ and $C$. The first operator $A$ produces 1% of 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 30% of the time and $C$ on the job for 20% of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by $A$?

$\frac{5}{34}$

The correct answer is Option (1) → $\frac{5}{34}$ ##

Let $A = \text{Event of choosing a defective item}$

$E_1 = \text{Item produced by A}$

$E_2 = \text{Item produced by B}$

$E_3 = \text{Item produced by C}$

$P(E_1) = \frac{50}{100} = \frac{1}{2}, \quad P(E_2) = \frac{30}{100} = \frac{3}{10}, \quad P(E_3) = \frac{20}{100} = \frac{1}{5}$

$P(A/E_1) = \frac{1}{100}, \quad P(A/E_2) = \frac{5}{100}, \quad P(A/E_3) = \frac{7}{100}$

Using Bayes' Theorem:

$P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)}$

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

$=\frac{\frac{1}{200}}{\frac{1}{200}+\frac{3}{200}+\frac{7}{500}}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{3}{2}+\frac{7}{5}}=\frac{\frac{1}{2}}{2+\frac{7}{5}}=\frac{\frac{1}{2}}{\frac{17}{5}}$

$P(E_1/A) =\frac{5}{34}$ = Probability that defective item was produced by A.