Question

A car manufacturing factory has two plants, $X$ and $Y$. Plant $X$ manufactures $70\%$ of cars and plant $Y$ manufactures $30\%.80\%$ of the cars at plant $X$ and $90\%$ of the cars at plant $Y$ are rated of standard quality. A car is chosen at random and is found to be of standard quality. The probability that is has come from plant $X$ is

• A. $\frac{56}{73}$
• B. $\frac{56}{84}$
• C. $\frac{56}{83}$
• D. $\frac{56}{79}$

Answer 1

Answer: (C)

Let $E$ be the event that the car is of standard quality. Let $B_1$ and $B_2$ be the events that the car is manufactured in plants $X$ and $Y$, respectively. Now,
$P\left(B_1\right)=\frac{70}{100}=\frac{7}{10},P\left(B_2\right)=\frac{30}{100}=\frac{3}{10}$
$P\left(E/B_1\right)$ - Probability that the car is manufactured in plant $X$, given that it is of standard quality
$=\frac{80}{100}=\frac{8}{10}\text{ and }P\left(E/B_2\right)=\frac{90}{100}=\frac{9}{10}$
$P\left(B_1/E\right)=$ Probability that the car is manufactured in plant $X$, given that it is of standard quality
$=\frac{P\left(B_1\right)\times P\left(E/B_1\right)}{P\left(B_1\right)\cdot P\left(E/B_1\right)+P\left(B_2\right)\cdot P\left(E/B_2\right)}$
$=\frac{\frac{7}{10}\times \frac{8}{10}}{\frac{7}{10}\times \frac{8}{10}+\frac{3}{10}\times \frac{9}{10}}=\frac{56}{83}$
Hence, the required probability is $\frac{56}{83}$.