Question

A n insurance company insured 2000 scoo terdrivers. 4000 car drivers and 6000 truck drivers , 7 he probeblitey of an accident involving a scooter. a and truck is $\frac{1}{100},\frac{3}{100}and\frac{3}{20}$ respectively. one of the insured persons meets with an accident. what is the probability that he is a scooter driver?

Answer 1

According to the problem,

Let$E_1,E_2\&E_3$ be the insured person of scooter, car and truck

Therefore,

$\begin{array}{c}P\left(E_1\right)=\frac{2000}{2000+4000+6000}=\frac{2000}{12000}\\ =\frac{1}{6}\\ P\left(E_2\right)=\frac{4000}{2000+4000+6000}=\frac{4000}{12000}\\ =\frac{1}{3}\\ P\left(E_3\right)=\frac{6000}{2000+4000+6000}=\frac{6000}{12000}\\ =\frac{1}{2}\end{array}$

Given $E_1$ be the person who insured with an accident.

Hence,

$\begin{array}{c}P\left(E/E_1\right)=\frac{1}{100}=0.01,P\left(E/E_2\right)=\frac{3}{100}0.03,P\left(E/E_3\right)=\frac{3}{20}=0.15\\ P\left(E_1/E\right)=\frac{P\left(E_1\right)P\left(E/E_1\right)}{P\left(E_1\right)P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)+P\left(E_3\right)P\left(E/E_3\right)}\\ =\frac{\left(\frac{1}{6}\right)\left(0.01\right)}{\left(\frac{1}{6}\right)\left(0.01\right)+\frac{1}{3}\left(0.03\right)+\frac{1}{2}\left(0.15\right)}\\ =\frac{\frac{1}{6}}{\frac{1}{6}+\frac{3}{3}+\frac{15}{2}}=\frac{1}{52}\end{array}$