Problem Statement
An insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probability of an accident involving a scooter, a car, and a truck are 0.01, 0.03, and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Verified Solution & Marking Scheme
Define Events and Prior Probabilities
Total insured drivers $= 2000 + 4000 + 6000 = 12000$.
Let:
- $E_1$: Person is a scooter driver $\implies P(E_1) = \frac{2000}{12000} = \frac{1}{6}$
- $E_2$: Person is a car driver $\implies P(E_2) = \frac{4000}{12000} = \frac{2}{6} = \frac{1}{3}$
- $E_3$: Person is a truck driver $\implies P(E_3) = \frac{6000}{12000} = \frac{3}{6} = \frac{1}{2}$
- $A$: Insured person meets with an accident.
State Likelihood Probabilities
$P(A|E_1) = 0.01 = \frac{1}{100}$
$P(A|E_2) = 0.03 = \frac{3}{100}$
$P(A|E_3) = 0.15 = \frac{15}{100}$
Apply Bayes' Theorem Formula
We need $P(E_1|A)$:
$P(E_1|A) = \frac{P(E_1) P(A|E_1)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + P(E_3) P(A|E_3)}$
Substitute Values and Evaluate
Denominator:
$P(A) = \frac{1}{6}\left(\frac{1}{100}\right) + \frac{2}{6}\left(\frac{3}{100}\right) + \frac{3}{6}\left(\frac{15}{100}\right)$
$= \frac{1 + 6 + 45}{600} = \frac{52}{600} = \frac{13}{150}$
Numerator:
$P(E_1) P(A|E_1) = \frac{1}{6}\left(\frac{1}{100}\right) = \frac{1}{600}$
$P(E_1|A) = \frac{1/600}{52/600} = \frac{1}{52}$.