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 $A$ were defective 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 it was produced by machine $B$?

Answer 1

Let $E_1$ 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.

$\therefore$ Probability of items produced by machine $A$, $P\left(E_1\right)=60$% $=\frac{3}{5}$

Probability of items produced by machine $B$, $P\left(E_2\right)=40$% $=\frac{2}{5}$

Probability that machine $A$ produced defective items, $P\left(X|E_1\right)=2$% $=\frac{2}{100}$

Probability that machine $B$ produced defective items, $P\left(X|E_2\right)=1$% $=\frac{1}{100}$

The probability that the randomly selected item was from machine $B$, given that it is defective, is given by $P\left(E_2|X\right)$.
By using Baye's theorem, we obtain

$P\left(E_2|X\right)=\frac{P\left(E_2\right)\cdot P\left(X|E_2\right)}{P\left(E_1\right)\cdot P\left(X|E_1\right)+P\left(E_2\right)\cdot P\left(X|E_2\right)}$

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

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

$=\frac{2}{8}$

$=\frac{1}{4}=0.25$