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

Question Number 31255 by Tip Top last updated on 04/Mar/18

2 n boys are randomly divided into two

subgroups containing n boys each . The

probability that the two tallest boys are

in different groups is

Answered by MJS last updated on 05/Mar/18

we have (2 n - 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

(\begin{matrix} n \\ 1 \end{matrix}) \times \frac{2}{2 n} \times \frac{2 n - 2}{2 n - 1} \times \frac{2 n - 3}{2 n - 2} \times \dots \times \frac{n}{n + 1} =

= \frac{2 n (2 n - 2)! / (n - 1)!}{(2 n)! / n!} =

= \frac{2 n (2 n - 2)! n!}{(2 n)! (n - 1)!} =

= \frac{n}{2 n - 1}

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