Question

The U.S. National Highway Traffic Safety Administration gathers data concerning the causes of highway crashes where at least one fatality has occurred. From the 1998 annual study, the following probabilities were determined (BAC is blood-alcohol level):

?(???=0|Crash with fatality)=0.625

?(??? is between .01 and .09|Crash with fatality)=0.302

?(??? is greater than .09|Crash with fatality)=0.069

Suppose over a certain stretch of highway during a 1-year period, the probability of being involved in a crash that results in at least one fatality is 0.028. It has been estimated that 14% of all drivers drive while their BAC is grater than .09. Determine the probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09).

Answer 1

Let event A: BAC=0

B: BAC is between 0.01 and 0.09

C: BAC is greater than 0.09

E: crash with fatality

So,

P(A|E)=0.625

P(B|E)=0.302

P(C|E)=0.069

given,

P(E)=0.028

P(BAC is greater than 0.09)=P(C)=0.14

Now

P(Crash with fatality| BAC is greater than 0.09)=P(E|C)

and

P(E|C)=P(E and E)/P(C)

now

we ahve

P(C|E)=0.069

=>P(C and E)/P(E)=0.069

=>P(C and E)=0.069*P(E)

=>P(C and E)=0.069*0.028

=>P(C and E)=0.001932

Now,

P(E|C)=P(E and E)/P(C)

=>P(E|C)=0.001932/0.14

=>P(E|C)=0.0138

probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09)=0.0138