In the final of a play competition for March 8, two plays made it to the final. In the first play, $n$ students from class 5A participated, and in the second play, $n$ students from class 5B participated. At the play, $2n$ mothers of all $2n$ students were present. The best play is chosen by the mothers' votes. It is known that each mother votes for the best play with a probability of $1 / 2$ and for the play in which her child participates with a probability of $1 / 2$.

a) Find the probability that the best play will win by a majority of votes.

b) The same question, if more than two classes made it to the final.

#
a) Let's call a mom confident if her child is playing in the best play. A confident mom will vote for the best play with probability 1. There are exactly $n$ confident moms, so the best play will get at least half of the votes - it will receive, at least, the votes of all confident moms. The only case when the best play does not win is a tie, when the worst play gets the votes of all uncertain moms. This is possible only if all uncertain moms vote for their own children. The probability of this event is $2^{-n}$. Therefore, the probability that more votes will be cast for the best play is $1-2^{-n}$.

b) Suppose that $m>2$ classes made it to the final, each with the same number $n$ of participants in the play. The best play will get at least $n$ votes - one vote from each confident mom. Each of the other plays will get no more than $n$ votes from all moms. If at least one uncertain mom votes for the best play, it wins. Therefore, the best play will not win only if all uncertain moms (there are exactly $(m-1) n$ of them) vote for their own children. The probability of this event is $2^{n-m n}$.

## Answer

a) $1-2^{-n}$;
b) $1-2^{n-m n}$,
where $m-$ is the number of classes.

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