Question

An insurance company insured 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 E₁, E₂ and E₃ denote the events that the vehicle is a scooter, a car and a truck, respectively.

Let A be the event that the vehicle meets with an accident.

It is given that there are 3000 scooters, 4000 cars and 5000 trucks.

Total number of vehicles = 3000 + 4000 + 5000 = 12000

P(E₁) = \[\frac{3000}{12000} = \frac{1}{4}\]

P(E₂) = \[\frac{4000}{12000} = \frac{1}{3}\]

P(E₃) = \[\frac{5000}{12000} = \frac{5}{12}\]

The probability that the vehicle, which meets with an accident, is a scooter is given by P (E₁/A).

\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 0 . 02 = \frac{2}{100}\]
\[P\left( A/ E_2 \right) = 0 . 03 = \frac{3}{100}\]
\[P\left( A/ E_3 \right) = 0 . 04 = \frac{4}{100}\]
\[\text{ Using Bayes' theorem, we get} \]
\[\left( i \right) \text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{4} \times \frac{2}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}\]
\[ = \frac{\frac{1}{2}}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{\frac{1}{2}}{\frac{3 + 6 + 10}{6}} = \frac{3}{19}\]
\[\left( ii \right) \text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_2 \right)P\left( A/ E_2 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{3} \times \frac{3}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}\]
\[ = \frac{1}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{1}{\frac{3 + 6 + 10}{6}} = \frac{6}{19}\]
\[\left( iii \right) \text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{5}{12} \times \frac{4}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}\]
\[ = \frac{\frac{5}{3}}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{\frac{5}{3}}{\frac{3 + 6 + 10}{6}} = \frac{10}{19}\]