Find the probability that a randomly selected person in a fatal car crash who was ejected from the vehicle was wearing a seat belt

Question

A study showed that 66% of occupants involved in a fatal car crash wore seat belts. Of those in a fatal car crash who wore seat belts, 16% were ejected from the vehicle. For those not wearing seat belts, 40% were ejected from the vehicle. Determine the following probabilities. Round your answers to four decimal places as needed. Q1 : Find the probability that a randomly selected person in a fatal car crash who was ejected from the vehicle was wearing a seat belt. Q2 : Find the probability that a randomly selected person in a fatal car crash who was not ejected from the vehicle was not wearing a seatbelt.

Answer 1

Applying Bayes' Theorem,

P(Seatbelt | Ejected) = (P(Ejected | Seatbelt)*P(Seatbelt)) / P(Ejected)

The denominator can be split conditionally, using the Law of Total Probability, based on whether or not the passenger was wearing a seatbelt, yielding

= (P(Ejected | Seatbelt)*P(Seatbelt)) / (P(Ejected | Seatbelt)P(Seatbelt) + P(Ejected | No Seatbelt)P(No Seatbelt))

A few things to notice: the numerator and the first term in the sum in the denominator are identical (P(Ejected | Seatbelt)P(Seatbelt)), as is common when applying Bayes' Theorem in this way. Secondly, I conditioned the way I did because it incorporates all of the known information.

The answer for part A is then

(.16 * .66) / (.16*.66 + .4 * (1 - .66)) = (.16 * .66) / (.16*.66 + .4 * .34)) = .4371

For part B, follow a similar approach:

P(No Seatbelt | Not Ejected) = (P(Not Ejected | No Seatbelt)*P(No Seatbelt)) / P(Not ejected)

= (P(Not Ejected | No Seatbelt)*P(No Seatbelt)) / (P(Not Ejected | No Seatbelt)*P(No Seatbelt) + P(Not Ejected | Seatbelt)*P(Seatbelt))

= ((1 - .4)*(1 - .66)) / ((1 - .4)*(1 - .66) + (1 - .16)*(.66))

= (.6 * .34) / (.6 * .34 + .84 * .66) = .2690