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# An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents 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? (Note, in 2014, this was modified to ask for the probability that he is a car driver also).

$\begin{array}{1 1} \large\frac{1}{52} \\ \large\frac{3}{52} \\ \large\frac{7}{52} \\ \large\frac{21}{52} \end{array}$

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Given the total number of drivers are as follows: 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total = 2000+4000+6000 = 12000.
Let $E_1$ be the event that the insured person is a scooter driver.
P ($E_1$) = $\large\frac{\text{Number of scooter drivers}}{\text{Total number of drivers}} = \frac{2000}{12000} = \frac{1}{6}$
Let $E_2$ be the event that the insured person is a car driver.
P ($E_2$) = $\large\frac{\text{Number of car drivers}}{\text{Total number of drivers}} = \frac{4000}{12000} = \frac{1}{3}$
Let $E_3$ be the event that the insured person is a truck driver.
P ($E_3$) = $\large\frac{\text{Number of truck drivers}}{\text{Total number of drivers}} = \frac{6000}{12000} = \frac{1}{2}$
Let A: event that the insured person met w an accident.
P (scooter driver met w/ an accident) = P (A|$E_1$) = 0.01 = $\large\frac{1}{100}$
P (car driver met w/ an accident) = P (A|$E_2$) = 0.03 = $\large\frac{3}{100}$
P (truck driver met w/ an accident) = P (A|$E_3$) = 0.15 = $\large\frac{15}{100}$
The probability that a driver is a scooter driver who met w/ an accident is given by $P(E_1/A)$.
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)+ P(E_3)P(A|E_3)}$
P ($E_1$|A) =$\large \frac{\frac{1}{6}.\frac{1}{100}}{\frac{1}{6} .\frac{1}{100}+\frac{1}{3}.\frac{3}{100}+\frac{1}{2}.\frac{15}{100}}$ =$\large\frac{1}{1+6+45}$ = $\large\frac{1}{52}$
Extra: We can easily show that P (that he is a car driver) $= \large\frac{6}{52}$
edited Mar 20, 2014