Let ${\mathit{E}}_1,{\mathit{E}}_2$ and ${\mathit{E}}_3$ be the events of a driver being a scooter driver, car driver and truck driver respectively. Let A be the event that the person meets with an accident.
There are 2000 insured scooter drivers, 4000 insured car drivers and 6000 insured truck drivers.
Total number of insured vehicle drivers = 2000 + 4000 + 6000 = 12000
∴ $\mathit{P}\left({\mathit{E}}_1\right)$ = $\frac{2000}{12000}$ = $\frac{1}{6}$ , $\mathit{P}\left({\mathit{E}}_2\right)$ = $\frac{4000}{12000}$ = $\frac{1}{3}$ , $\mathit{P}\left({\mathit{E}}_3\right)$ = $\frac{6000}{12000}$ = $\frac{1}{2}$
Also, we have:
P (A|${\mathit{E}}_1$) = 0.01 = $\frac{1}{100}$
P (A|${\mathit{E}}_2$) = 0.03 = $\frac{3}{100}$
P (A|${\mathit{E}}_3$) = 0.15 = $\frac{15}{100}$
Now, the probability that the insured person who meets with an accident is a scooter driver is P (A|${\mathit{E}}_1$).
Using Bayes’ theorem, we obtain:
P (${\mathit{E}}_1$|A) =
= $\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}$
= $\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}$
= $\frac{1}{6}\times \frac{6}{52}$
= $\frac{1}{52}$