Question

A manufacturer has three machine operators $A,B$ and $C$. The first operator $A$ produces $1\%$ defective items, whereas the other two operators $B$ and $C$ produce $5\%$ and $7\%$ defective items respectively. $A$ is on the job for $50\%$ of the time, $B$ is on the job for $30\%$ of the time and $C$ is on the job for $20\%$ of the time. A defective item is produced, what is the probability that it was produced by $A$?

Answer 1

Let $E_1,E_2$ and $E_3$ be the respective events of the time consumed by machines $A,B$ and $C$ for the job.

$P\left(E_1\right)=50$% $=\frac{50}{100}=\frac{1}{2}$

$P\left(E_2\right)=30$% $=\frac{30}{100}=\frac{3}{10}$

$P\left(E_3\right)=20$% $=\frac{20}{100}=\frac{1}{5}$

Let $X$ be the event of producing defective items.
$P\left(X|E_1\right)=1$% $=\frac{1}{100}$

$P\left(X|E_2\right)=5$% $=\frac{5}{100}$

$P\left(X|E_3\right)=7$% $=\frac{7}{100}$

The probability that the defective item was produced by $A$ is given by $P\left(E_1|A\right)$.
By using Baye's theorem, we obtain

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

$=\frac{\frac{1}{2}\cdot \frac{1}{100}}{\frac{1}{2}\cdot \frac{1}{100}+\frac{3}{10}\cdot \frac{5}{100}+\frac{1}{5}\cdot \frac{7}{100}}$

$=\frac{\frac{1}{100}\cdot \frac{1}{2}}{\frac{1}{100}\left(\frac{1}{2}+\frac{3}{2}+\frac{7}{5}\right)}$

$=\frac{\frac{1}{2}}{\frac{17}{5}}$

$=\frac{5}{34}=0.14$