Question

An insurance company issued 3000 scooters, 4000 cars and 5000 trucks. The probabilities of the accident involving a scooter, a car and a truck are 0.02, 0.03 and 0.04 respectively. One of the insured vehicles meet with an accident. Find the probability that it is a (i) scooter (ii) car (iii) truck.

Answer 1

Let $E_1,E_2$ and$E_3$ be the events of the insured vehicle i.e., scooter, car and truck

$P\left(E_1\right)=\frac{3000}{3000+4000+5000}=\frac{3}{12}$

$P\left(E_2\right)=\frac{4000}{3000+4000+5000}=\frac{4}{12}$

$P\left(E_3\right)=\frac{5000}{3000+4000+5000}=\frac{5}{12}$

Let $A$ be the event that vehicle meets an accident

$\left(A/E_1\right)=0.02$

$P\left(A/E_2\right)=0.03$

$P\left(A/E_3\right)=0.04$

By Baye's rule

$\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)+P\left(E_3\right).P\left(A/E_3\right)}=\frac{\frac{3}{12}\times 0.02}{\frac{3}{12}\times 0.02+\frac{4}{12}\times 0.03+\frac{5}{12}\times 0.04}=\frac{3}{19}$

$P\left(E_2/A\right)=\frac{6}{19}$ and similarly $P\left(E_3/A\right)=\frac{10}{29}$