Question

Alcohol is a major factor in traffic accidents. In 2019, the U.S. Department of Transportation, National Highway Traffic Safety Administration, stated that among age 21 to 24 year olds, 29% of drivers involved in fatal crashes were driving under the influence (of alcohol). Suppose three fatal car accidents are randomly selected and examined. (a) (3 points) What is the probability that the first accident selected had an individual driving under the influence of alcohol but the next two were not? (b) (4 points) What is the probability that at least one of the accidents involved a driver under the influence of alcohol? Is this unusual? Explain.

Answer 1

P(driving under influence) = 0.29

P(not driving under influence) = 1-0.29 = 0.71

a.

probability that the first accident selected had an individual driving under the influence of alcohol but the next two were not

P(first accident selected had an individual driving under the influence of alcohol but the next two were not)

= P(driving under influence)*P(not driving under influence)*P(not driving under influence)

= 0.29*0.71*0.71

= 0.146189

b.

probability that at least one of the accidents involved a driver under the influence of alcohol

P(atleast one driving under influence) = 1 - P(none driving under influence)

= 1 - P(not driving under influence)^3

= 1 - 0.71^3

= 0.642089

P(atleast one driving under influence) = 0.642089

this is not unusual as probability is more than 50%

(please UPVOTE)