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. What is the probability that he is a scooter driver?

Answer 1

There are 2000 scooter drivers, 4000 car drivers and 6000 truck drivers.

Total number of drivers =2000+4000+6000=12000

Let $E_1$ the event that insured person is a scooter driver, $E_2$ the event that insured person is a car driver and $E_3$ the event that insured person is a truck driver.

Then, $E_1,E_2,E_3$ are mutually exclusive and exhaustive events. Moreover,

P$E_1=\frac{Numberofscooterdrivers}{Totalnumberofdrivers}=\frac{2000}{12000}=\frac{1}{6}$

P$E_2=\frac{Numberofcardrivers}{Totalnumberofdrivers}=\frac{4000}{12000}=\frac{1}{3}$

and P$E_1=\frac{Numberoftruckdrivers}{Totalnumberofdrivers}=\frac{6000}{12000}=\frac{1}{2}$

Let E: the events that insured person meets with an accident,

P$\left(\frac{E}{E_1}\right)$ = P (scooter driver met with an accident)=0.001=$\frac{1}{100}$

P$\left(\frac{E}{E_2}\right)$ = P (car driver met with an accident)=0.03=$\frac{3}{100}$

P$\left(\frac{E}{E_3}\right)$ = P (scooter driver met with an accident)=0.15=$\frac{15}{100}$

The probability that the driver is a scooter driver, given he met with an accident, is given by P$\left(\frac{E}{E_1}\right)$.

By using Baye's theorem, we obtain

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

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