**Question:**

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

**Solution:**

Let E1, E2, and E3 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

$P\left(E_{1}\right)=P$ (driver is a scooter driver) $=\frac{2000}{12000}=\frac{1}{6}$

$P\left(E_{2}\right)=P$ (driver is a car driver) $=\frac{4000}{12000}=\frac{1}{3}$

$P\left(E_{3}\right)=P$ (driver is a truck driver) $=\frac{6000}{12000}=\frac{1}{2}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}($ scooter driver met with an accident $)=0.01=\frac{1}{100}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}($ car driver met with an accident $)=0.03=\frac{3}{100}$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{3}\right)=\mathrm{P}($ truck driver met with an accident $)=0.15=\frac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P\left(E_{1} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(A \mid E_{3}\right)}$

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

$=\frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{100}\left(\frac{1}{6}+1+\frac{15}{2}\right)}$

$=\frac{\frac{1}{6}}{\frac{104}{12}}$

$=\frac{1}{6} \times \frac{12}{104}$

$=\frac{1}{52}$

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