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

$P\left(E_2|A\right)=\frac{P\left(E_2\right)P\left(A|E_2\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)+P\left(E_3\right)P\left(A|E_3\right)}$

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

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