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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
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 E1 = the item is manufactured by the operator A, E2 = the item is manufactured by the operator B, E3 = the item is manufactured by the operator C and A = the item is defective
Now P (E1) = $\frac{{50}}{{100}}$, P(E2) = $\frac{{30}}{{100}}$, P(E3) = $\frac{{20}}{{100}}$
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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