An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers.

Solution:

Let E1, E2, and E3 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

$P\left(E_{1}\right)=P$ (driver is a scooter driver) $=\frac{2000}{12000}=\frac{1}{6}$

$P\left(E_{2}\right)=P$ (driver is a car driver) $=\frac{4000}{12000}=\frac{1}{3}$

$P\left(E_{3}\right)=P$ (driver is a truck driver) $=\frac{6000}{12000}=\frac{1}{2}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}($ scooter driver met with an accident $)=0.01=\frac{1}{100}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}($ car driver met with an accident $)=0.03=\frac{3}{100}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{3}\right)=\mathrm{P}($ truck driver met with an accident $)=0.15=\frac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P\left(E_{1} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(A \mid E_{3}\right)}$

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

$=\frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{100}\left(\frac{1}{6}+1+\frac{15}{2}\right)}$

$=\frac{\frac{1}{6}}{\frac{104}{12}}$

$=\frac{1}{6} \times \frac{12}{104}$

$=\frac{1}{52}$