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 on the job for 30% of the time and C 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 E₁, E₂ and E₃ be the time taken by machine operators A, B, and C, respectively.

Let X be the event of producing defective items.

$$\begin{gathered} \therefore P\left(E_1\right)=50\%=\frac{1}{2} \\ P\left(E_2\right)=30\%=\frac{3}{10} \\ P\left(E_3\right)=20\%=\frac{1}{5} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=1\%=\frac{1}{100} \\ P\left(A/E_2\right)=5\%=\frac{5}{100} \\ P\left(A/E_2\right)=7\%=\frac{7}{100} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)++P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{100}}}{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{100}}+{\displaystyle \frac{3}{10}}\times {\displaystyle \frac{5}{100}}+{\displaystyle \frac{1}{5}}\times {\displaystyle \frac{7}{100}}} \\ =\frac{5}{34} \end{gathered}$$