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)=0.631P(BAC=0|Crash with fatality)=0.631

?(??? is between .01 and .09|Crash with fatality)=0.325P(BAC is between .01 and .09|Crash with fatality)=0.325

?(??? is greater than .09|Crash with fatality)=0.098P(BAC is greater than .09|Crash with fatality)=0.098

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 0.022. 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).

Probability =  

Answer 1

Answer:

Here from above P(crash with fatality|BAC is greater then 0.09)

=P(crash with fatality)*P(BAC is greater then 0.09|crash with fatality)/P(BAC is greater then 0.09)

=0.022 * 0.098/ 0.14 = 0.0154

please revert

[OR}

Solution:

We are given here that:
P( BAC = 0 | crash with fatality ) = 0.631
P( 0.01 < BAC < 0.09 | crash with fatality) = 0.325
P( BAC > 0.09 | crash with fatality ) = 0.098

P( crash with fatality ) = 0.022

P( BAC > 0.09) = 0.14

Using Bayes theorem, we have here:
P( BAC > 0.09 | crash with fatality )P( crash with fatality ) = P( crash with fatality | BAC > 0.09)P(BAC > 0.09)

Putting all the above values, we get here:
0.098 *0.022 = P( crash with fatality | BAC > 0.09)*0.14
P( crash with fatality | BAC > 0.09) = 0.098*0.022/ 0.14 = 0.0154

Therefore 0.0154 is the required probability here.

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