A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as 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 was produced by A?
Let E1, E2, and E3 be the respective events of the time consumed by machines A, B, and C for the job.
$\mathrm{P}\left(\mathrm{E}_{1}\right)=50 \%=\frac{50}{100}=\frac{1}{2}$
$\mathrm{P}\left(\mathrm{E}_{2}\right)=30 \%=\frac{30}{100}=\frac{3}{10}$
$\mathrm{P}\left(\mathrm{E}_{3}\right)=20 \%=\frac{20}{100}=\frac{1}{5}$
Let X be the event of producing defective items.
$\mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)=1 \%=\frac{1}{100}$
$\mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)=5 \%=\frac{5}{100}$
$\mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{3}\right)=7 \%=\frac{7}{100}$
The probability that the defective item was produced by $A$ is given by $P\left(E_{1} \mid A\right)$.
By using Bayes’ theorem, we obtain
$P\left(E_{1} \mid X\right)=\frac{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(X \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(X \mid E_{3}\right)}$
$=\frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{2} \cdot \frac{1}{100}+\frac{3}{10} \cdot \frac{5}{100}+\frac{1}{5} \cdot \frac{7}{100}}$
$=\frac{\frac{1}{100} \cdot \frac{1}{2}}{\frac{1}{100}\left(\frac{1}{2}+\frac{3}{2}+\frac{7}{5}\right)}$
$=\frac{\frac{1}{2}}{\frac{17}{5}}$
$=\frac{5}{34}$