An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents 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 ?

Open in App

Solution

Let $E_{1}, E_{2}$ , and $E_{3}$ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let $A$ be the event that the person meets with an accident.
There are $2000$ scooter drivers, $4000$ car drivers, and $6000$ truck drivers.
Total number of drivers $= 2000 + 4000 + 6000 = 12000$
$P (E_{1}) = P (d r i v e r i s a s c o o t e r d r i v e r) = \frac{2000}{12000} = \frac{1}{6}$
$P (E_{2}) = P (d r i v e r i s a c a r d r i v e r) = \frac{4000}{12000} = \frac{1}{3}$
$P (E_{3}) = P (d r i v e r i s a t r u c k d r i v e r) = \frac{6000}{12000} = \frac{1}{2}$
$P (A | E_{1}) = P (s c o o t e r d r i v e r m e t w i t h a n a c c i d e n t) = 0.01 = \frac{1}{100}$
$P (A | E_{2}) = P (c a r d r i v e r m e t w i t h a n a c c i d e n t) = 0.03 = \frac{3}{100}$
$P (A | E_{3}) = P (t r u c k d r i v e r m e t w i t h a n a c c i d e n t) = 0.15 = \frac{15}{100}$
The probability that the driver is a scooter driver, given that he met with an accident, is given by $P (E_{1} | A)$ .
By using Baye's theorem, we obtain
$P (E_{1} | A) = \frac{P (E_{1}) \cdot P (A | E_{1})}{P (E_{1}) \cdot P (A | E_{1}) + P (E_{2}) \cdot P (A | E_{2}) + P (E_{3}) \cdot P (A | E_{3})}$
$= \frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{6} \cdot \frac{1}{100} + \frac{1}{3} \cdot \frac{3}{100} + \frac{1}{2} \cdot \frac{15}{100}}$
$= \frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{100} (\frac{1}{6} + 1 + \frac{15}{2})}$
$= \frac{\frac{1}{6}}{\frac{52}{6}}$
$= \frac{1}{6} \times \frac{12}{104}$
$= \frac{1}{52} = 0.019$

Suggest Corrections

Similar questions

2000

4000

6000

0.01, 0.03

0.15

\frac{p}{q}

q - p

\frac{1}{100}, \frac{3}{100} a n d \frac{3}{20}

Join BYJU'S Learning Program