• anonymous
probability help please .....During random roadside stops, 5% of drivers sow traces of alcohol, and 10% of drivers are not wearing seat belts. Assume these two infractions are independent from each other. If a police officer stops 20 cars what is the probability that exactly 4 of the drivers show traces of alcohol?
  • Stacey Warren - Expert
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  • chestercat
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  • Directrix
Because the infractions are indepedent and the question asks only about traces of alcohol, I am assuming that seat belts is not of concern here. Using binomial probability, the probability that out of 20 randomly chosen stops, exactly 4 stops will involve traces of alcohol is the following: C(20, 4) * (.05)^4 * (.95)^0 where .95 is the probability that a stop will not involve alcohol. C(20,4) means 20, choose 4 which is 20!/(4!*16!) = 4845 P( 4 alcohol stops) = C(20, 4) * (.05)^4 * (.95)^0 becomes 4845 * (.05)^4 * 1= 0.03028125 or .03 approximately. Please check my work.

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