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): P(BAC
P(????????????=0|Crash with fatality)=0.625

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

P(???????????? 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

Answer:

probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09) = 0.001932

Step-by-step explanation:

P(BAC=0|Crash with fatality)=0.625

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

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

Let the event of BAC = 0 be X

Let the event of BAC between 0.01 and 0.09 be Y

Let the event of BAC greater than 0.09 be Z

Let the event of a crash with at least one fatality = C

P(X|C) = 0.625

P(Y|C) = 0.302

P(Z|C) = 0.069

P(C) = 0.028

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

But note that the conditional probability of probability that a driver is intoxicated (BAC greater than 0.09) given that there was a crash that involved at least a fatality is given by

P(Z|C) = P(Z n C)/P(C)

P(Z n C) = P(Z|C) × P(C) = 0.069 × 0.028 = 0.001932

Hope this Helps!!!