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# hard probability problem

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Let \(S=\{1, 2, 3,\dots, n\}.\) Three subsets  \(A, B, C\)  of \(S\) are chosen at random. Find the probability that \(A\subseteq B\subseteq C.\)

\(A\subseteq B\subseteq C\)just means all elements of  \(A\) are elements of \(B\) and all elements of \(B\) are elements of \(C.\)

Sep 12, 2021

#1
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There are 2^n ways of choosing C, then 2^(n - 1) ways of choosing B, then 2^(n - 2) ways of choosing A, so the probability is 2^n*2^(n - 1)*2^(n - 2)/(2^n*2^n*2^n) = 1/8.

Sep 12, 2021
#2
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B is not strictly one element less than C.

Guest Sep 13, 2021