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?

Solution

Let E₁, E₂ and E₃ be the events of a driver being a scooter driver, car driver and truck driver respectively. Let A be the event that the person meets with an accident.

There are 2000 insured scooter drivers, 4000 insured car drivers and 6000 insured truck drivers.

Total number of insured scooter drivers vehicle drivers = 2000 + 4000 + 6000 = 12000

$∴ P (E_{1}) = \frac{2000}{12000} = \frac{1}{6}, P (E_{2}) = \frac{4000}{12000} = \frac{1}{3}, P (E 3) = \frac{6000}{12000} = \frac{1}{2},$

Also, we have:

$P (A Ι E_{1}) = 0.01 = \frac{1}{100} P (A Ι E_{2}) = 0.03 = \frac{3}{100} P (A Ι E_{3}) = 0.15 = \frac{15}{100}$

Now, the probability that the insured person who meets with an accident is a scooter driver is $P (E_{1} Ι A) .$

Using Baye's theorem, we obtain:

$P (E_{1} Ι A) = \frac{P (E_{1}) x P (A Ι E_{1})}{P (E_{1}) x P (A Ι E_{1}) + P (E_{2}) x P (A Ι E_{2}) + P (E_{3}) x P (A Ι E_{3})} = \frac{\frac{1}{6} x \frac{1}{100}}{\frac{1}{6} x \frac{1}{100} + \frac{1}{3} x \frac{3}{100} + \frac{1}{2} x \frac{15}{100}} = \frac{\frac{1}{6} x \frac{1}{100}}{\frac{1}{6} + 1 + \frac{15}{2}} = \frac{1}{6} x \frac{6}{52} = \frac{1}{52}$

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