An insurance company insured 200 scooter drivers, 400 car drivers and 600 truck drivers. The probability of an accident involving scooter driver, car driver and truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured person meets an accident, then the probability that he is a truck driver is:-
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{45}{52} \)
Let \( E_1,E_2,E_3 \) be the events that the person is a scooter, car or truck driver, and A the event of an accident. Out of 1200 insured people:
\( \displaystyle P(E_1)=\frac{200}{1200}=\frac{1}{6} \)
\( \displaystyle P(E_2)=\frac{400}{1200}=\frac{1}{3} \)
\( \displaystyle P(E_3)=\frac{600}{1200}=\frac{1}{2} \)
By Bayes' theorem:
\( \displaystyle P(E_3|A)=\frac{P(E_3)P(A|E_3)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)} \)
\( \displaystyle =\frac{\frac{1}{2}(0.15)}{\frac{1}{6}(0.01)+\frac{1}{3}(0.03)+\frac{1}{2}(0.15)} \)
Multiplying the numerator and the denominator by 1200:
\( \displaystyle =\frac{600(0.15)}{200(0.01)+400(0.03)+600(0.15)} \)
\( \displaystyle =\frac{90}{2+12+90} \)
\( \displaystyle =\frac{90}{104}=\frac{45}{52} \)
Shortcut: By the ratio method, the expected numbers of accidents are 2 (scooter), 12 (car) and 90 (truck). The required probability is \( \frac{90}{2+12+90}=\frac{45}{52} \).