A manufacturer has three machine operators A, B and C.

Solution:

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

$P\left(E_{1} \mid X\right)=\frac{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(X \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(X \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(X \mid E_{3}\right)}$

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

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

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

$=\frac{5}{34}$