JEE Advanced 2016 · Paper 1 · Mathematics Q40

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MathematicsSingle correct+3 / -1Question 40 of 54

A computer producing factory has only two plants $T_1$ and $T_2$ . Plant $T_1$ produces 20% and plant $T_2$ produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that

$P$ (computer turns out to be defective given that it is produced in plant $T_1$ ) $=10P$ (computer turns out to be defective given that it is produced in plant $T_2$ ),

where $P(E)$ denotes the probability of an event $E$ . A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

\dfrac{36}{73}

\dfrac{47}{79}

\dfrac{78}{93}

\dfrac{75}{83}

Answer and worked solution

Official answer: C

Defect rates. Let $P(D|T_2)=p$ , so $P(D|T_1)=10p$ . Then
$0.07=0.2(10p)+0.8p=2.8p\implies p=0.025.$

Not defective.

$P(\bar D|T_1)=0.75$ and $P(\bar D|T_2)=0.975$ .
$P(\bar D)=0.2(0.75)+0.8(0.975)=0.15+0.78=0.93$ .

Bayes.
$P(T_2|\bar D)=\frac{0.78}{0.93}=\frac{78}{93}.$

(C)

Topics:Bayes' theorem Conditional probability

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JEE Advanced 2016 · Paper 1, Mathematics question 40