Question 579317: On New Years eve,
The probability of driving while intoxicated is .32
The probability of having a driving accident is .09
What is the probability of a person having a driving accident while intoxicated?
Found 2 solutions by edjones, Theo:
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! Since the chance of having an accident while driving intoxicated are greatly increased the probability cannot be calculated from the statistics given.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! assuming the probability of getting into an accident while intoxicated is the same as the probability of getting into an accident while sober, then the probability of a driver being in an accident and being intoxicated at the same time is .09 * .32 = .0288.
this means that 2.88% of the time a driver will be in an accident and be intoxicated at the same time.
if the probabiltity of getting into an accident while intoxicated was 5 times the national overall average, and the overall probability was still .09, then the probability of getting into an accident while intoxicated and while sober would become:
5x * .32 + x * .68 = .09
solve for x to get:
x = .09/2.28 = .0394736842
5x = .1973684211
probability of getting into an accident while intoxicated becomes .1973684211 and the probability of getting into an accident while sober becomes .0394736842.
the overall rate is still .09 because .1973684211 * .32 + .0394736842 * .68 = .09
what does all this mean?
assume 1000 drivers
assume the probability of getting into an accident is .09 for intoxicated drivers and .09 for sober drivers.
.09 *.32 * 1000 equals 28.8 drivers get in an accident while intoxicated and .09 * .68 * 1000 equals 61.2 drivers get in an accident while sober.
the overall number is 28.8 + 61.2 = 90 drivers get in an accident, whether intoxicated or sober.
90/1000 = .09 which is the overall rate.
now assume the probability of getting into an accident is .1973684211 while intoxicated and .0394736842 while sober.
.973684211 * .32 * 1000 equals 63.15789474 drivers get in an accident while intoxicated and .0394736842 * .68 * 1000 equals 26.84210526 drivers get in an accident while sober.
the overall number is 63.15789474 + 26.84210526 = 90 drivers get in an accident, whether intoxicated or sober.
90/100 = .09 which is the same overall rate.
only the distribution is different.
your answer depends heavily on the probability of getting into an accident when intoxicated and when sober.
the probability is different.
i don't really know how different, but it is different.
i used 5 times the probability just to show you the impact of it being different assuming that the probability was 5 times the probability of getting into an accident while intoxicated over getting into an accident while sober.
the actual statistics may be higher or they may be lower.
your question doesn't take any of this into account, so the answer to your question is conditional on the fact that the probability of getting into an accident while intoxicated is the same as the probability of getting into an accident while sober.
in that case, the answer is .32 * .09 = .0288
this is the probability of getting into an accident and being intoxicated at the same time.
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