Question

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.

Answer 1

Probability that he is a scooter driver =scooter drivers/all drivers

2000/12000

1/6

So that answers is 1/6

Answer 2

Answer:

= $\frac{1}{52}$

Step-by-step explanation:

Let P(A) = P(scooter) = $\frac{2000}{12000} = \frac{1}{6}$

      P(B) = P(car) = $\frac{4000}{12000} = \frac{1}{3}$

      P(C) = P (truck) = $\frac{6000}{12000} = \frac{1}{2}$

Let E = Event that person meets with accident

Then,       $P(\frac{E}{A}) = \frac{1}{100} , P(\frac{E}{B}) = \frac{3}{100} , P (\frac{E}{C}) = \frac{15}{100}$

Required Probability,

Baye's Theorem

      P(A) . P(E/A) / P(A) . P(E/A) + P(B) . P(E/B) + P(C) . P(E/C)

 = 1/6 × 1/100 / 1/6 × 1/100 + 1/3 × 3/100 + 1/2 × 15/100

    = 1/6 / 1/6 + 1 + 15/2

   = 1/6 / 1 + 6 + 45 / 6

    = $\frac{1}{52}$

Good luck !!