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ltruong2
Average Member

USA
18 Posts

 Posted - 10/10/2011 :  03:22:52 Would someone please show me how to find P(M1), P(F2), and P(M3) from the problem belowProblem: A company employs six male executives and three female executives. Three executives are randomly selected, one at a time, from the nine executives; the first one selected is sent to China, the second one is sent to New York, and third one is sent to France. Using the general multiplication rule, find the probability that a male is sent to China, a female is sent to New York, and a male is sent to France. Express your answer as a reduced fraction. Hint: Write the probability as P(M1F2M3), then apply the rule.Here is how I approach the solution:1. the Sample space S = 9(8)(7) = 504 ways to choose the 3 executives randomly2. P(M1F2M3) = P(M1)*p(F2|M1)*P(M3|M1F2)3. But I am not sure how to find P(M1) or P(F2) or P(M3)

royhaas
Moderator

USA
3059 Posts

 Posted - 10/12/2011 :  15:18:22 Calcuate this in order. How many males are there? What is the probability that a male goes to China? How many females are there? What is the probability that a female goes to NY? How many males are left after one goes to China?

ltruong2
Average Member

USA
18 Posts

 Posted - 10/13/2011 :  01:55:08 Thanks for the help.I think I got it already.
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