Question

An insurance company insured $1800$ car drivers, $1500$ bike drivers and $5000$ truck drivers. The probability of a car, a bike and a truck driver meeting with an accident is $0.02$, $0.03$ and $0.06$ respectively. If one of the insured person meets with an accident, then find the probability that he is a bike driver.

• A. $\frac{15}{127}$
• B. $\frac{15}{129}$
• C. $\frac{14}{127}$
• D. $\frac{13}{129}$

Answer 1

Answer: (A)

Let $E_1$, $E_2$ and $A$ be the events defined as follows :
$E_1 =$ insured person is a car driver
$E_2 =$ insured person is a bike driver
$E_3 =$ insured person is a truck driver
$A =$ person meets with an accident
Total number of insured persons
$= 1800+ 1500 + 5000 = 8300$
$\therefore P\left(E_{1}\right) = \frac{1800}{8300} = \frac{18}{83}$,

$P\left(E_{2} \right) = \frac{1500}{8300} = \frac{15}{83}$,

$P\left(E_{3}\right) = \frac{5000}{8300} = \frac{50}{83}$

$P\left(A|E_{1}\right) = P$(a car driver meeting with an accident) $= 0.02$
$P\left(A|E_{2}\right) = 0.03$, $P\left(A|E_{3}\right) = 0.06$
We want to find $P\left(E_{2}|A\right)$
By Bayes' theorem, we have

$P\left(E_{2}|A\right) = \frac{P\left(E_{2}\right)P\left(A |E_{2}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)+ P\left(E_{3}\right)P\left(A|E_{3}\right)}$

$= \frac{\frac{15}{83}\times0.03}{\frac{18}{83}\times 0.02+\frac{15}{83}\times 0.03+\frac{50}{83}\times 0.06}$

$= \frac{0.45}{3.81} = \frac{15}{127}$