Question

In a bolt factory, three machines $A,B$ and $C$ manufacture $25$%, $35$% and $40$% of the total production respectively. Of their respective outputs, $5$%, $4$% and $2$% are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine $C$.

Answer 1

Let $E_1$: the event that bolt is produced by machine $A$.
$E_2$: the event that bolt is produced by machine $B$.
and $E_3$: the event that bolt is produced by machine $C$.
Here $E_1,E_2$ and $E_2$ are mutually exclusive and exhaustive events.
We have,
$P\left(E_1\right)=25$% $=\frac{25}{100}$

$P\left(E_2\right)=35$% $=\frac{35}{100}$$P\left(E_3\right)=40$% $=\frac{40}{100}$

Let $E$: the event that bolt chosen is found to be defective.

$\therefore P\left(\frac{E}{E_1}\right)=5$% =$\frac{5}{100},P\left(\frac{E}{E_2}\right)=4$% $=\frac{4}{100}$ and $P\left(\frac{E}{E_3}\right)=2$% $=\frac{2}{100}$$P$ (defective bolt is produced by machine $C$)

$P\left(\frac{E_3}{E}\right)=\frac{P\left(E_3\right)\cdot P\left(\frac{E_3}{E}\right)}{P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right)\cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right)\cdot P\left(\frac{E}{E_3}\right)}$

$=\frac{\frac{40}{100}\times \frac{2}{100}}{\frac{25}{100}\times \frac{5}{100}+\frac{35}{100}\times \frac{4}{100}+\frac{40}{100}\times \frac{2}{100}}$

$=\frac{80}{125+140+80}$

$=\frac{80}{345}=0.23$