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Probability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
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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