Question

A recent highway safety study found that in 65% of all accidents a driver was wearing a seatbelt. Accident reports indicated that 83% of those drivers escaped serious injury (defined as hospitalization or death), but only 49% of the non-belted drivers were so fortunate. Find the probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured. Show your work (if using notations, make sure to identify them). (Round your answer to 2 places after the decimal point).

Answer 1

Answer:

The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E) = 0.76

Step-by-step explanation:

Probability of wearing a seatbelt in an accident = P(B) = 65% = 0.65

Probability of not wearing a seatbelt in an accident = P(B') = 1 - 0.65 = 0.35

Probability of escaping hospitalization and/or death given that one is wearing a seatbelt = P(E|B) = 83% = 0.83

Probability of escaping hospitalization and/or death given that one isn't wearing a seatbelt = P(E|B') = 0.49

Find the probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)

The probability of P(X|Y) is given mathematically as P(X n Y)/P(Y)

P(B|E) = P(B n E)/P(E)

But P(E) is unknown at the moment.

But P(E) = P(B n E) + P(B' n E) mathematically,.

P(B n E) can be obtained using P(E|B) and P(B)

P(E|B) = P(B n E)/P(B)

P(B n E) = P(E|B) × P(B) = 0.83 × 0.65 = 0.5395

And

P(B' n E) can be obtained using P(E|B') and P(B')

P(E|B') = P(B' n E)/P(B')

P(B' n E) = P(E|B') × P(B') = 0.49 × 0.35 = 0.1715

P(E) = P(B n E) + P(B' n E) = 0.5395 + 0.1715 = 0.711

The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)

P(B|E) = P(B n E)/P(E) = 0.5395/0.711 = 0.76