Question

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident for them are 0.01, 0.03 and 0.15, respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver or a car driver?

Answer 1

Let S , C , T be the events that a scooter driver, a car driver and a truck driver are insured.
And let A be the event that an insured person meets with an accident.
Here,

\[P\left( S \right) = \frac{2000}{12000} = \frac{1}{6} , P\left( C \right) = \frac{4000}{12000} = \frac{1}{3} , P\left( T \right) = \frac{6000}{12000} = \frac{1}{2}\]

\[\text { And } P\left( AS \right) = 0 . 01 , P\left( AC \right) = 0 . 03 , P\left( AT \right) = 0 . 15\]

So, using the theorem of total probability, we get:

\[P\left( A \right) = P\left( S \right) \times P\left( AS \right) + P\left( C \right) \times P\left( AC \right) + P\left( T \right) \times P\left( AT \right)\]

\[ = \frac{1}{6} \times 0 . 01 + \frac{1}{3} \times 0 . 03 + \frac{1}{2} \times 0 . 15\]

\[ = \frac{13}{150}\]

Now, using Bayes' theorem, we get:

\[P\left( \left( C or S \right) A \right) = \frac{P(C) \times P(AC) + P(S) \times P(AS)}{P(A)} = \frac{\frac{1}{3} \times 0 . 03 + \frac{1}{6} \times 0 . 01}{\frac{13}{150}}\]

\[ = \frac{7}{52}\]