A manufacturer has three machine operators A, B and C. The first … - Vidyadip

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 $A$?

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Answer

Let $E_1, E_2$ and $E_3$ 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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