Question

An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01,0.03$ and $0.15$ respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver?

Answer 1

Given the total number of drivers are as follows: 2000 scooter drivers, 4000 car drivers and 6000 truck driver. $Total=2000+4000+6000=12000$

Let $E_1$ be the event that the insured person is a scooter driver

Then $P\left(E_1\right)=\frac{Numberofscooterdrivers}{Totalnumber}=\frac{2000}{12000}=\frac{1}{6}$

Let $E_2$ be the event that the insured person is a car driver

Then $P\left(E_2\right)=\frac{Numberofcardrivers}{Totalnumber}=\frac{4000}{12000}=\frac{1}{3}$

Let $E_3$ be the event that the insured person is a truck driver

Then $P\left(E_3\right)=\frac{Numberoftruckdrivers}{Totalnumber}=\frac{6000}{12000}=\frac{1}{2}$

Let $A$ be the event that the insured person met with an accident.

P (scooter driver met with an accident) = $P\left(A|E_1\right)=0.01=1/100$

P (car driver met with an accident) = $P\left(A|E_2\right)=0.03=3/100$

P (truck driver met with an accident) = $P\left(A|E_2\right)=0.15=15/100$

The probability that a driver is a scooter driver who met w/ an accident is given by $P\left(E_1|A\right)$.

We can use Baye's theorem, according to which

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

$=\frac{\frac{1}{6}\cdot \frac{1}{100}}{\frac{1}{6}\cdot \frac{1}{100}+\frac{1}{3}\cdot \frac{3}{100}+\frac{1}{2}\cdot \frac{15}{100}}=\frac{1}{1+6+45}=\frac{1}{52}$

Hence, the probability is $\frac{1}{52}$.