Question

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 $T _{1}$ ) $=10 P$ (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 -

• A. $\frac{36}{73}$
• B. $\frac{47}{79}$
• C. $\frac{78}{93}$
• D. $\frac{75}{83}$

Answer 1

Answer: (C)

Given $P\left(\frac{D}{T_{1}}\right)=10 P\left(\frac{D}{T_{2}}\right)$
Now $P(D)=P\left(T_{1}\right) P\left(\frac{D}{T_{1}}\right)+P\left(T_{2}\right) P\left(\frac{D}{T_{2}}\right)$
$\frac{7}{100}=\frac{20}{100} \times 10 P\left(\frac{D}{T_{2}}\right)+\frac{80}{100} P \left(\frac{D}{T_{2}}\right)$
$\therefore P\left(\frac{D}{T_{2}}\right)=\frac{7}{280}, P\left(\frac{D}{T_{1}}\right)=\frac{70}{280}$
Now $P\left(\frac{T_{2}}{\bar{D}}\right)=\frac{P\left(T_{2}\right) P\left(\frac{\bar{D}}{T_{2}}\right)}{P\left(T_{1}\right) P\left(\frac{\bar{D}}{T_{1}}\right)+P\left(T_{2}\right) P\left(\frac{\bar{D}}{T_{2}}\right)}$
$=\frac{\frac{80}{100} \times \frac{273}{280}}{\frac{20}{100} \times \frac{210}{280}+\frac{80}{100} \times \frac{273}{280}}=\frac{78}{93}$