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

What is the probability a driver will be taken the hospital this year (as a result of an accident)? Given that a driver wears a seatbelt, what is the probability he/she will be taken to the hospital this year?

Answer 1

Probability that the drivers are safe, P(S) = 0.6

Probability that the drivers are unsafe, P(NS) = 1 - 0.6 = 0.4

Probability of safe drivers who wear seatbelts , P(SBS) = 0.25*P(S) = 0.25*0.6 = 0.15

Probability of unsafe drivers wear seatbelts , P(SBNS) = 0.10 * P(NS) = 0.1*0.4 = 0.04

Therefore, Total Probability of wearing seatbelts ,P(S) = P(SBS) + P(SBNS) = 0.15 + 0.04 = 0.19.

Total Probability of not wearing seatbelts ,P(NwS) = 1 - 0.19 = 0.81.

Probability of safe drivers experience an accident in a year, P(AS) = 0.1 * P(S) = 0.1 * 0.6 = 0.06

Probability of unsafe drivers experience an accident in a year , P(ANS)= 0.2 * P(NS) = 0.2 * 0.4 = 0.08.

Therefore, Total Probability of Accidents , P(A) = P(AS) + P(ANS) = 0.06 + 0.08 = 0.14.

Probability that driver will be hosptialised in an accident, given that he wears seat belt, is given by,

P(H | S) = 0.01

Probability that they will be hosptialised, given that they met with an accident while not wearing a seatbelt is given by,

P(H | NwS) = 0.2

Therefore, Total Probability that driver will be hospitalized ,

P(H) = P(H | S) * P(S) + P(H | NwS) * P(Nws) = 0.01 * 0.19 + 0.2 * 0.81 = 0.1639

Now, this hospitiization will happen , only when we have an accident..

Therefore, probability a driver will be taken the hospital this year = 0.1639*P(A) = 0.1639*0.14 = 0.0229 = 2.29%

Now, when a driver wears a seatbelt , the probability he/she will be taken to the hospital this year is given by,

P = P(H | S) * P(S) /  P(H | S) * P(S) + P(H | NwS) * P(Nws) = (0.01 * 0.19) / (0.01 * 0.19 + 0.2 * 0.81) = 0.0115 = 1.15%