Question 1114854: Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrive first and Keith last? Find the probability that Jean will arrive first and Keith will arrive last.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time.
In how many ways can they arrive? ----- 6!
In how many ways can Jean arrive first and Keith last?----1*4!*1 = 24
Find the probability that Jean will arrive first and Keith will arrive last.
----- 24/6! = 0.0333..
Cheers,
Stan H.
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