Question

In any given year, an insurance company believes the following:

-0.6 of drivers are safe.

-0.25 of safe drivers wear seatbelts

-0.10 of unsafe drivers wear seatbelts

-0.10 of safe drivers experience an accident in a year

-0.20 of unsafe drivers experience an accident in a year

-Given a driver experiences an accident, the driver has probability 0.01 they will be taken to the hospital if they were wearing a seatbelt

-Given a driver experiences an accident, the driver has probability 0.2 they will be taken to the hospital if they were not wearing a seatbelt

Given that a driver is taken to the hospital, what is the probability they were not wearing a seatbelt?

Answer 1

Let A denote the event that the driver is safe

B denote the event that the driver is wearing a seatbelt

C denote the event that the driver experience an accident

D denote the event that driver will be taken to the hospital given he/she experiences an accident

P(A) = 0.6 , P(A') = 0.4

P(B | A) = 0.25, P(B | A') = 0.10

P(C | A) = 0.10, P(C | A') = 0.20

P(D | B) = 0.01, P(D | B') = 0.2

The required probability = P(B' | D)

P(B) = P(B | A)*P(A) + P(B | A')*P(A') = 0.25*0.6 + 0.1*0.4 = 0.19

P(D) = P(D | B)*P(B) + P(D | B')*P(B') = 0.01*0.19 + 0.2*0.81 = 0.1639

P(B | D) = P(D | B)*P(B)/P(D)

= 0.01*0.19/0.1639 = 0.0116

Thus, the required probability = P(B' | D)

= 1 - P(B | D)

= 0.9884