Question

An insurance company insured $2000$ scooter drivers, $4000$ car drivers, and $6000$ truck drivers. The probability of accidents are $0.01,0.03$ and $0.15$, respectively. One of the insured persons meets with an accident. The probability that he is a scooter driver is $\frac{p}{q}$ then $q-p$ is

Answer 1

Let $E_1,E_2$ and $E_3$ be the respective events that the driver is a scooter driver, a car driver and a truck driver.
Let $A$ be the event that the person meets with an accident.
There are $2000$ scooter drivers, $4000$ car drivers, and $6000$ truck drivers.
Total number of drivers $=2000+4000+6000=12000$
$P\left(E_1\right)=P\left(\text{driver is a scooter driver}\right)=\frac{2000}{12000}=\frac{1}{6}$

$P\left(E_2\right)=P\left(\text{driver is a car driver}\right)=\frac{4000}{12000}=\frac{1}{3}$

$P\left(E_3\right)=P\left(\text{driver is a truck driver}\right)=\frac{6000}{12000}=\frac{1}{2}$

$P\left(A/E_1\right)=P\left(\text{scooter driver meets with an accident}\right)=0.01=\frac{1}{100}$

$P\left(A/E_2\right)=P\left(\text{car driver meets with an accident}\right)=0.03=\frac{3}{100}$

$P\left(A/E_3\right)=P\left(\text{truck driver meets with an accident}\right)=0.15=\frac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P\left(E_1/A\right)$.
By using Baye's theorem, we obtain
$P\left(E_1/A\right)=\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)+P\left(E_3\right).P\left(A/E_3\right)}$
$=\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}=\frac{1}{52}$
$\frac{1}{52}=\frac{p}{q}$
$\Rightarrow q-p=52-1=51$