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, where as the other two operators $B$ and $C$ produce $5\%$ and $7\%$ defective items respectively. A is on the job for $50\%$ of the time, $B$ is 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 $E_1 =$ the item is manufactured by the operator $A, E_2 =$ the item is manufactured by the operator $B, E_3 =$ the item is manufactured by the operator $C$ and $A =$ the item is defective
Now $P (E_1) = \frac{{50}}{{100}}$, $P(E_2) = \frac{{30}}{{100}}$, $P(E_3) = \frac{{20}}{{100}}$
Now $P\left( {A|{E_1}} \right)$ = P(item drawn is manufactured by operator A) = $\frac{1}{{100}}$
Similarly, $P\left( {A|{E_2}} \right)$ = $\frac{5}{{100}}$ and $P\left( {A|{E_3}} \right)$ = $\frac{7}{{100}}$
Now Required probability = Probability that the item is manufactured by operator. A given that the item drawn is defective
$P(\frac {{E_1}} {A})$ = $\frac{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A|{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A|{E_3}} \right)}}$= $\frac{{\frac{{50}}{{100}} \times \frac{1}{{100}}}}{{\frac{{50}}{{100}} \times \frac{1}{{100}} + \frac{{30}}{{100}} \times \frac{5}{{100}} \times \frac{7}{{100}}}}$
= $\frac{{50}}{{50 + 150 + 140}} = \frac{5}{{34}}$

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