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=0|Crash with fatality)=0.611 P(BAC is between .01 and .09|Crash with fatality)=0.347 P(BAC is greater than .09|Crash with fatality)=0.054 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 0.027. It has been estimated that 15% 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

- data concerning the Solutions * it is given that the teallic safety administeation gathers causes of highway " i hichway clashes where at least on fatality has occurred. * The probabilites given ale → PCBA C=0)) reaish with fatality) - 0.611 eash with fatality) = 0.347 → PCBAC is between 0.01 € 0.09 Ceash with fatality) = 0.054 → P(BAC is greater than 0.09 ceash with 1. let us considee. * The event A represents that BAC =0 * " " B " BAC is between 0.01€0.09 *, C BAL is greater than 0.09 * " " t represents that ceash with fatality. So, the given probabilites are pCAle] - 0.611 PCBle] = 0.347 P[cle] = 0.054 * It is given that the probability g being involved in a ceash that results in at least the fatality is 0.007, over a certain Stretch of highway during I year per p (leash with Jatulity) - 0.007 le >) PCE) -0.007

* It has been estimated that 1st of the drivers on this highway drive while their BAC is greater than 0.09 PCBAC is geeater than 0.09)-0.15 => P[C] = 0.15 * we have to find the probability of a ceash with at least ! fatality ij a deives deives while legally intoxicated We have to find pleash with fatality I BAC is geeater than 0.097 which means P[Elc) =) P[6/c] - PC4EC) PCC] ist we have to find eLE&c] We have PCCIE) - 0.054 -> p (EEC) -0.054 Pre] -> PCE@C) - 0.054 P CE] >> PCE &C] -10.054)0.027) > p (EEC)0.001458 .: the required probability is obtained below PCeIc) - PCEE C) PEC] = 0.01458 / 0.15 0.0972

Thus p(EIC)= 0.0972 . .. The probability of aceash with at least I fatality if adeives deives while legally intoxicated is 0.0972