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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability3 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 $C$ is on the job for $20\%$ of the time. A defective item is produced, what is the probability that it was produced by $A?$
✓

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
$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{50}{100},\ \text{P}(\text{E}_2)=\frac{30}{100},\ \text{P}(\text{E}_3)=\frac{20}{100}$
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}$

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