Question

The members of a consulting firm rent cars from three rental agencies: $50\%$ from agency $X,30\%$ from agency $Y$ and $20\%$ from agency $Z$.From past experience it is known that $9\%$ of the cars from agency $X$ need a service and tuning before renting , $12\%$ of the cars agency $Y$ need a service and tuning before rating and $10\%$ of the cars from agency $Z$ need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency $Z$ is not to be blamed.

Answer 1

Let us take

A: car needs service

$E_1$: car is rented from x

$E_2$: car is rented from y

$E_3$: car is rented from z

We have to find the probability that car is choosen from z, if car needs service$P\left(E_3/A\right)$

To find not choosen probability

$P\left(E_1\right)=50\%=\frac{50}{100}$

$P\left(E_2\right)=30\%=\frac{30}{100}$

$P\left(E_3\right)=20\%=\frac{20}{100}$

$P\left(A/E_1\right)=$probability that car needs service

$=9\%=\frac{9}{100}$

$P\left(A/E_2\right)=12\%=\frac{12}{100}$

$P\left(A/E_3\right)=10\%=\frac{10}{100}$

$P\left(E_3/A\right)=\frac{P\left(E_3\right)\cdot P\left(A/E_3\right)}{P\left(E_1\right)\cdot P\left(A/E_1\right)+P\left(E_2\right)\cdot P\left(A/E_2\right)+P\left(E_3\right)\cdot P\left(A/E_3\right)}$

$=\frac{\frac{20}{100}\times \frac{10}{100}}{\frac{50}{100}\times \frac{9}{100}+\frac{30}{100}\times \frac{12}{100}+\frac{20}{100}\times \frac{10}{100}}$

$=\frac{200}{450+360+200}$

$=\frac{200}{1010}$

$=\frac{20}{101}$

According to question we have to find car is not choosen from agency$=4$, if it needs service.

So, $\Rightarrow 1-P\left(E_3/A\right)$

$=1-\frac{20}{101}$

$=\frac{101-20}{101}$

$=\frac{81}{101}$

Required answer.