An insurance company insures 4000 two-wheeler drivers, 5000 car drivers and 1000 truck drivers. The probabilities that a two-wheeler driver, a car driver and a truck driver meet with an accident in a year are 0.02, 0.01 and 0.05 respectively. One of the insured drivers meets with an accident. The probability that the driver is a two-wheeler driver is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{9}\)
There are 10000 drivers in all. Let \(E_1, E_2, E_3\) be the events that the driver drives a two-wheeler, a car and a truck, and let A be the event of an accident.
\(P(E_1) = 0.4,\quad P(E_2) = 0.5,\quad P(E_3) = 0.1\)
\(P(E_1)P(A|E_1) = 0.4\times 0.02 = 0.008\)
\(P(E_2)P(A|E_2) = 0.5\times 0.01 = 0.005\)
\(P(E_3)P(A|E_3) = 0.1\times 0.05 = 0.005\)
By the theorem of total probability:
\(P(A) = 0.008 + 0.005 + 0.005 = 0.018\)
By Bayes’ theorem:
\( \displaystyle P(E_1|A) = \frac{0.008}{0.018} = \frac{4}{9} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{5}{18}\) — Incorrect. This is \(\frac{0.005}{0.018}\), the chance that the driver is a car driver (the same value holds for a truck driver).
Option (3) \(\displaystyle\frac{2}{5}\) — Incorrect. This is the share of two-wheeler drivers among all insured drivers, before the accident is known.
Option (4) \(\displaystyle\frac{1}{4}\) — Incorrect. This is \(\frac{0.02}{0.02+0.01+0.05}\), which uses the accident rates but ignores how many drivers of each kind there are.