Question

In a factory, there are two machines $A$ and $B$ producing bulbs. Their contributions are respectively $60\%$ and $40\%$ of the total production. It is found that $1\%$ and $3\%$ of the items produced by these machines are defective. An item randomly chosen from a day's production was found to be defective. Find the probability that the defective item was produced by machine $B$.

Answer 1

Let $E_1$ and $E_2$ be the events that the items produced by machine $A$ and $B$ respectively.

Then $P\left(E_1\right)=\frac{60}{100}=\frac{3}{5}$ and $P\left(E_2\right)=\frac{40}{100}=\frac{2}{5}$

Let $C$ be the event that we choose a defective item at random.

Then, $P\left(\text{The defective item came from machine}A\right)=P\left(C|E_1\right)=1\%=\frac{1}{100}$

$P\left(\text{The defective item came from machine}B\right)=P\left(C|E_2\right)=3\%=\frac{3}{100}$

To find the probability that the defective item was produced by machine $B$, we need to find $P\left(E_2|C\right)$

So, we use Baye's formula and then we have

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

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

$=\frac{6}{3+6}=\frac{6}{9}=\frac{2}{3}$

Hence, the probability that the defective item was produced by machine $B$ is $\frac{2}{3}$.