Question

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 and 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. $\frac{19}{52}$

Answer 1

The correct option is A $\frac{1}{52}$

Solution:

Let $P\left(A\right)=P\left(scooter\right)=\frac{2000}{12000}=\frac{1}{6}$

$P\left(B\right)=P\left(car\right)=\frac{4000}{12000}=\frac{1}{3}$

and $P\left(C\right)=P\left(truck\right)=\frac{6000}{12000}=\frac{1}{2}$

Let $E=$ Event the person meets with an accident.

Then, $P\left(E/A\right)=\frac{1}{100},P\left(E/B\right)=\frac{3}{100},P\left(E/C\right)=\frac{15}{100}$

Now, $P\left(A/E\right)=\frac{P\left(A\right).P\left(E/A\right)}{P\left(A\right).P\left(E/A\right)+P\left(B\right).P\left(E/B\right)+P\left(C\right).P\left(E/C\right)}$

$=\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}$

$=\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}$

$=\frac{1}{52}$

Hence, A is the correct option.