One of the insured persons meets with an accident. What is the probability that the person is a scooter driver? The image displays the Question i am confused with. Help please > 102. An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are $0.01$, $0.03$, $0.15$ respectively. One of the insured persons meets with an accident. The probability that the person is a scooter driver is > (a) $\frac{1}{52}$ > (b) $\frac{3}{52}$ > (c) $\frac{15}{52}$ > (d) $\frac{19}{52}$ >

Let the probabilities of the insured person be a scooter driver be $P(S)$, be a car driver be $P(C)$ and a truck driver be $P(T)$. Given that $P(S) = \frac{2000}{12000} = \frac{1}{6}$, $P(C) = \frac{1}{3}$ and $P(T) = \frac{1}{2}$. If $A$ is the accident event, $P(A|S) = 0.01, P(A|C) = 0.03, P(A|T) = 0.15$. We want $P(S|A)$. This by Bayes's formula is given by \begin{align*} P(S|A) &= \frac{P(A|S)P(S)}{P(A)} \\\ &= \frac{P(A|S)P(S)}{P(A|S)P(S)+P(A|C)P(C)+P(A|T)P(T)}\\\ &= \frac{0.01 \times \frac{1}{6}}{0.01 \times \frac{1}{6}+0.03 \times \frac{1}{3}+0.15 \times \frac{1}{2}}\\\ &=\frac{1}{52} \end{align*}