Q69 / 180 · KCET 2021 · Mathematics

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

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

Solution

Let $E=$ the event that the car is of standard quảlity. Let $A_{1}=$ the event that the car is manufactured in plant $X$. and $A_{2}=$ the event that the car is manufactured in plant $Y$. Now, $P\left(A_{1}\right)=\frac{70}{100}=\frac{7}{10}, P\left(A_{2}\right)=\frac{30}{100}=\frac{3}{10}$ $P\left(\frac{E}{A_{1}}\right)=$ Probability that a standard quality car is manufactured in plant $X=\frac{80}{100}=\frac{8}{10}$ $P\left(\frac{E}{A_{2}}\right)=$ Probability that a standard quality car is manu factured in blant $Y=\frac{90}{100}=\frac{9}{10}$ $P\left(\frac{A_{1}}{E}\right)=$ Probability that a standard quality car has come from plant $X$ $\begin{aligned} =& \frac{P\left(A_{1}\right) \times P\left(\frac{E}{A_{1}}\right)}{P\left(A_{1}\right) \cdot P\left(\frac{E}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{E}{A_{2}}\right)} \\ =& \frac{\frac{7}{10} \times \frac{8}{10}}{\frac{7}{10} \times \frac{8}{10}+\frac{3}{10} \times \frac{9}{10}} \\ =& \frac{7 \times 8}{7 \times 8+3 \times 9} \\ =& \frac{56}{56+27}=\frac{56}{83} \end{aligned}$ Hence, the required probability is $\frac{56}{83}$.

Hence, the correct answer is option: 'C'.

Relevant Questions