In a factory which manufactures bolts, machines A, B and C manufacture respectively 25\%, 35\% and 40\% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

$\frac{28}{69}$

The correct answer is Option (2) → $\frac{28}{69}$ ##

Let events $B_1$, $B_2$, $B_3$ be the following:

$B_1$ : the bolt is manufactured by machine A

$B_2$ : the bolt is manufactured by machine B

$B_3$ : the bolt is manufactured by machine C

Clearly, $B_1, B_2, B_3$ are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space.

Let the event $E$ be `the bolt is defective'.

The event $E$ occurs with $B_1$ or with $B_2$ or with $B_3$. Given that,

$P(B_1) = 25\% = 0.25, \quad P(B_2) = 0.35 \text{ and } P(B_3) = 0.40$

Again $P(E|B_1) = $ Probability that the bolt drawn is defective given that it is manufactured by machine A $= 5\% = 0.05$

Similarly, $P(E|B_2) = 0.04, \quad P(E|B_3) = 0.02$.

Hence, by Bayes' Theorem, we have

$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2) + P(B_3)P(E|B_3)}$

$= \frac{0.35 \times 0.04}{0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02}$

$= \frac{0.0140}{0.0345} = \frac{28}{69}$