Question

An insurance company insured 2000 scooter drivers, 4000 car drivers & 6000 truck drivers. The probabilities of their accident are 0.1, 0.3, and 0.2 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer 1

Let $E_1$ , $E_2$  and $E_3$  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

$\therefore P\left(E_{1}\right)=\frac{2000}{12000}=\frac{1}{6}, P\left(E_{2}\right)=\frac{4000}{12000}=\frac{1}{3}, P\left(E_{3}\right)=\frac{6000}{12000}=\frac{1}{2}$

Also, we have:

 $P\left(A \mid E_{1}\right)=0.01=\frac{1}{100}$

$\begin{aligned} &P\left(A \mid E_{2}\right)=0.03=\frac{3}{100} \\ &P\left(A \mid E_{3}\right)=0.15=\frac{15}{100} \end{aligned}$

 Now, the probability that the insured person who meets with an accident is a scooter driver is $P\left(E_{1} \mid A\right).$

Using Bayes' theorem, we obtain:

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

$\begin{aligned} &=\frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100}+\frac{1}{3} \times \frac{3}{100}+\frac{1}{2} \times \frac{15}{100}} \\ &=\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}} \\ &=\frac{1}{6} \times \frac{6}{52} \\ &=\frac{1}{52} \end{aligned}$