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