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

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 is 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
$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{50}{100},\ \text{P}(\text{E}_2)=\frac{30}{100},\ \text{P}(\text{E}_3)=\frac{20}{100}$
Now $\text{P}(\text{A}|\text{E}_1)$ = P(item drawn is manufactured by operator A) $=\frac{1}{100}$
Similarly, $\text{P}(\text{A}|\text{E}_2)=\frac{5}{100}\ \text{and}\ \text{P}(\text{A}|\text{E}_3)=\frac{7}{100}$
Now Required probability = Probability that the item is manufactured by operator A given that the item drawn is defective
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)}$
$ =\frac{\frac{50}{100}\times\frac{1}{100}}{\frac{50}{100}\times\frac{1}{100}+\frac{30}{100}\times\frac{5}{100}+\frac{20}{100}\times\frac{7}{100}}=\frac{50}{50+150+40}=\frac{5}{34}$