**Sol:**

E_{1} and E_{2} 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.

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

The Probability of items produced by machine\( A, P(E_{2}) = 40% = \frac{2}{5}\)

The probability that machine A produced defective items,\(P(X \mid E_{1}) = 2% = \frac{2}{100}\)

The probability that machine B produced defective items,\(P(X \mid E_{2}) = 1% = \frac{1}{100}\)

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

By using Baye’s theorem, we obtain:

\(P (E_{2}\mid X) = \frac{P(E_{2})*P(X\mid E_{2})}{P(E_{1}) * P(X \mid E_{1}) * P(X \mid E_{2})}\\\Rightarrow \frac{\frac{2}{5} * \frac{1}{100}}{\frac{3}{5}*\frac{2}{100} + \frac{2}{5} * \frac{1}{100}}\\\Rightarrow \frac{\frac{2}{500}}{\frac{5}{500} + \frac{2}{500}}\\\Rightarrow \frac{2}{8}\\\Rightarrow \frac{1}{4} = 0.25\)Therefore, the probability that it was produced by machine B is 0.25.

Explore more such questions and answers at BYJUâ€™S.