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Question
An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
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Solution
Let A, E1 and E2 denote the events that the vehicle meets the accident, is a scooter and is a motorcycle, respectively.
\[\therefore P\left( E_1 \right) = \frac{2000}{5000} = 0 . 4 \]
\[ P\left( E_2 \right) = \frac{3000}{5000} = 0 . 6\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 0 . 01\]
\[P\left( A/ E_2 \right) = 0 . 02\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_2 \right)P\left( A/ E_2 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{0 . 6 \times 0 . 02}{0 . 6 \times 0 . 02 + 0 . 4 \times 0 . 01}\]
\[ = \frac{0 . 012}{0 . 012 + 0 . 004} = \frac{0 . 012}{0 . 016} = \frac{3}{4}\]
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