# A factory has two machines A and B.

Question:

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. 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 was produced by machine B?

Solution:

Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

$\therefore$ Probability of items produced by machine $A, P\left(E_{1}\right)=60 \%=\frac{3}{5}$

Probability of items produced by machine $\mathrm{B}, \mathrm{P}\left(\mathrm{E}_{2}\right)=40 \%=\frac{2}{5}$

Probability that machine A produced defective items, $P\left(X \mid E_{1}\right)=2 \%=\frac{2}{100}$

Probability that machine $B$ produced defective items, $P\left(X \mid E_{2}\right)=1 \%=\frac{1}{100}$

The probability that the randomly selected item was from machine $B$, given that it is defective, is given by $P\left(E_{2} \mid X\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{2} \mid X\right)=\frac{P\left(E_{2}\right) \cdot P\left(X \mid E_{2}\right)}{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(X \mid E_{2}\right)}$

$=\frac{\frac{2}{5} \cdot \frac{1}{100}}{\frac{3}{5} \cdot \frac{2}{100}+\frac{2}{5} \cdot \frac{1}{100}}$

$=\frac{\frac{2}{500}}{\frac{6}{500}+\frac{2}{500}}$

$=\frac{2}{8}$

$=\frac{1}{4}$