Question

In a factory machine $A$ produce $30\%$ of the total output, machine $B$ produced $25\%$ and the machine $C$ produced the remaining output. If defective items produced by machine $A,B$ and $C$ are $1\%,12\%,2\%$ respectively. Three machines working together produced $10000$ item in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine $B$.

Answer 1

Let $E_1,E_2,E_3$ be events as define below:

$E_1=$Item is produced by machine $A$,

$E_2=$Item is produced by machine $B$

$E_3=$Item is produced by machine $C$ and

$A=$That event that the item is defective.

It is given that, $P\left(E_1\right)=\frac{30}{100},P\left(E_2\right)=\frac{25}{100},P\left(E_3\right)=\frac{45}{100}$,

$P\left(A/E_1\right)=\frac{1}{100}$

$P\left(A/E_2\right)=\frac{1.2}{100}$

$P\left(A/E_3\right)=\frac{2}{100}$

Required probability $=P\left(E_2/A\right)=\frac{P\left(E_2\right)P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)+P\left(E_3\right)P\left(A/E_3\right)}$

$=\frac{\frac{25}{100}\times \frac{1.2}{100}}{\frac{30}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{1.2}{100}+\frac{2}{100}\times \frac{45}{100}}$

$=\frac{30}{30+30+90}=\frac{1}{5}$