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 $1\%$ 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 $A$?

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Answer

Let $E_1, E_2,$ and $E_3$ be the respective events of the time consumed by machine $A, B,$ and $C$ for the job.
$\text{P}(\text{E}_1)=50\%=\frac{50}{100}=\frac{1}{2}$
$\text{P}(\text{E}_2)=30\%=\frac{30}{100}=\frac{3}{10}$
$\text{P}(\text{E}_3)=20\%=\frac{20}{100}=\frac{1}{5}$
Let $X$ be the event of producing defective items.
$\text{P}(\text{X}|\text{E}_1)=1\%=\frac{1}{100}$
$\text{P}(\text{X}|\text{E}_2)=5\%=\frac{5}{100}$
$\text{P}(\text{X}|\text{E}_3)=7\%=\frac{7}{100}$
The probability that the defective item was produced by $A$ is given by $P(E_1|A).$
By using Bayes' theorem, we obtain
$\text{P}(\text{E}_1|\text{X})=\frac{\text{P}\text{E}_1\text{P}(\text{X}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{X}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{X}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{X}|\text{E}_3)}$
$=\frac{\frac{1}{2}\times\frac{1}{100}}{\frac{1}{2}\times\frac{1}{100}+\frac{3}{10}\times\frac{5}{100}+\frac{1}{5}\times\frac{7}{100}}$
$=\frac{\frac{1}{100}\times\frac{1}{2}}{\frac{1}{100}\Big(\frac{1}{2}+\frac{3}{2}+\frac{7}{5}\Big)}$
$=\frac{\frac{1}{2}}{\frac{17}{5}}$
$=\frac{5}{34}$

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