In a factory, machine A produces 30% of total output, machine B produces 25% and the machine C produces the remaining output. The defective items produced by machines A, B and C are 1%, 1.2%, 2%, respectively. An item is picked at random from a day's output and found to be defective. Find the probability that it was produced by machine B?

$\frac{1}{5}$

The correct answer is Option (2) → $\frac{1}{5}$ ##

Let $E_1 = \text{Choosing machine A}$

$E_2 = \text{Choosing machine B}$

$E_3 = \text{Choosing machine C}$

$A = \text{Producing a defective output}$

Given,

$P(E_1) = 30\% = \frac{30}{100} = 0.3$

$P(E_2) = 25\% = \frac{25}{100} = 0.25$

$P(E_3) = [100 - (30 + 25)]\% = 45\% = \frac{45}{100} = 0.45$

and $P\left( \frac{A}{E_1} \right) = P(\text{Producing defective output from machine } A) $

$= 1\% = \frac{1}{100} = 0.01$

$P\left( \frac{A}{E_2} \right) = P(\text{Producing defective output from machine } B)$

$ = 1.2\% = \frac{1.2}{100} = 0.012$

$P\left( \frac{A}{E_3} \right) = P(\text{Producing defective output from machine } C)$

$ = 2\% = \frac{2}{100} = 0.02$

Required probability: $P\left( \frac{E_2}{A} \right)$

$ = P(\text{The found defective item is produced by machine } B)$

Using Bayes' theorem,

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

$= \frac{0.25 \times 0.012}{(0.3 \times 0.01) + (0.25 \times 0.012) + (0.45 \times 0.02)}$

$= \frac{0.003}{0.003 + 0.003 + 0.009} = \frac{0.003}{0.015} = \frac{3}{15} = \frac{1}{5}$

Thus, the required probability is \( \frac{1}{5} \).