Question
A manufacturer has three machine operators A, B and C. The first operator A produces 1% of defective items, whereas the other two operators B and C produces 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job for 20% of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by A?

Answer

Let E1, E2 and E3 be the event that machine is operated by A, B, and C respectively.
Let A be the event of producing defective items.
$\therefore\text{P}(\text{E}_1)=50\%=\frac{1}{2}$
$\text{P}(\text{E}_2)=30\%=\frac{3}{10}$
$\text{P}(\text{E}_3)=20\%=\frac{1}{5}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1\%=\frac{1}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=5\%=\frac{5}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=7\%=\frac{7}{100}$
Using Bayes' theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\big(\frac{\text{A}}{\text{E}_1}\big)}{\text{P}(\text{E}_1)\text{P}\big(\frac{\text{A}}{\text{E}_1}\big)+\text{P}(\text{E}_2)\text{P}\big(\frac{\text{A}}{\text{E}_2}\big)+\text{P}(\text{E}_3)\text{P}\big(\frac{\text{A}}{\text{E}_3}\big)}$
$=\frac{\frac{1}{2}\times\frac{1}{100}}{\frac{1}{2}\times\frac{1}{100}+\frac{3}{10}\times\frac{5}{100}+\frac{1}{5}\times\frac{7}{100}}$
$=\frac{5}{34}$

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