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 $T_1$ ) $=10P$ (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 -

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

Answer 1

Answer: (C)

Given $P\left(\frac{D}{T_1}\right)=10P\left(\frac{D}{T_2}\right)$
Now $P(D)=P\left(T_1\right)P\left(\frac{D}{T_1}\right)+P\left(T_2\right)P\left(\frac{D}{T_2}\right)$
$\frac{7}{100}=\frac{20}{100}\times 10P\left(\frac{D}{T_2}\right)+\frac{80}{100}P\left(\frac{D}{T_2}\right)$
$\therefore P\left(\frac{D}{T_2}\right)=\frac{7}{280},P\left(\frac{D}{T_1}\right)=\frac{70}{280}$
Now $P\left(\frac{T_2}{\bar{D}}\right)=\frac{P\left(T_2\right)P\left(\frac{\bar{D}}{T_2}\right)}{P\left(T_1\right)P\left(\frac{\bar{D}}{T_1}\right)+P\left(T_2\right)P\left(\frac{\bar{D}}{T_2}\right)}$
$=\frac{\frac{80}{100}\times \frac{273}{280}}{\frac{20}{100}\times \frac{210}{280}+\frac{80}{100}\times \frac{273}{280}}=\frac{78}{93}$