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2n-boys-are-randomly-divided-into-two-subgroups-containing-n-boys-each-The-probability-that-the-two-tallest-boys-are-in-different-groups-is-

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Question Number 31255 by Tip Top last updated on 04/Mar/18
2n boys are randomly divided into two subgroups containing n boys each. The probability that the two tallest boys are in different groups is
2nboysarerandomlydividedintotwosubgroupscontainingnboyseach.Theprobabilitythatthetwotallestboysareindifferentgroupsis
Answered by MJS last updated on 05/Mar/18
we have (2n−2) black marbles and 2 red ones we draw n marbles without returning we search the probability for exactly 1 red and (n−1) black marbles ((n),(1) )×(2/(2n))×((2n−2)/(2n−1))×((2n−3)/(2n−2))×...×(n/(n+1))= =((2n(2n−2)!/(n−1)!)/((2n)!/n!))= =((2n(2n−2)!n!)/((2n)!(n−1)!))= =(n/(2n−1))
wehave(2n−2)blackmarblesand2redoneswedrawnmarbleswithoutreturningwesearchtheprobabilityforexactly1redand(n−1)blackmarbles(n1)×22n×2n−22n−1×2n−32n−2×…×nn+1==2n(2n−2)!/(n−1)!(2n)!/n!==2n(2n−2)!n!(2n)!(n−1)!==n2n−1

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