A factory has two machines A and B.

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

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

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

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

$=\frac{2}{8}$

$=\frac{1}{4}$