Suppose that $A$ and $B$ are independent of each other, equally likely to be any of the $2^n$ subsets of $\{1, \ldots , n \}$,I want to find

a) $P(A\subset B)$.

b) $P(A \cap B) = \varnothing$.


To answer this question, is the knowledge of $\sigma$ algebra necessary?

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    $\begingroup$ You do not need to consider the sigma-algebra explicitly: you can just treat the status of each integer $i$'s membership of $A$ and $B$ to be independent of the other integers' statuses, so take the answers when $n=1$ and raise them to the power of $n$ $\endgroup$ – Henry Jan 13 '16 at 16:54
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    $\begingroup$ can you make the question title more specific? nearly any probability question could have this title. $\endgroup$ – Charlie Parker Jan 13 '16 at 19:59

For part $(a)$.

$\text{pr}\left(A \subset B\right) = \sum_{\bar{B}} \text{pr}\left(A \subset B | B=\bar{B}\right) \text{pr}\left(B = \bar{B}\right)$.

Because each set $B$ is equally likely: $\text{pr}\left(B = \bar{B}\right) = 2^{-n}$.

Because $B$ and $A$ are independent $\text{pr}\left(A \subset B | B=\bar{B}\right)=\text{pr}\left(A \subset \bar{B}\right)$.

For each $\bar{B}$, how many subsets does $\bar{B}$ have compared the total number of subsets ($2^{n}$)? If $\bar{B}$ has $k$ elements, that means it has $2^{k}$ subsets so: $\text{pr}\left(A \subset \bar{B}\right) = 2^{k} / 2^{n}$.

How many sets $\bar{B}$ have $k$ elements? There are $n$ choose $k$ subsets. So:

$\text{pr}\left(A \subset B\right) = 2^{-n} \sum_{k=1}^{n} {n \choose k} 2^{k-n}$.

A similar strategy should work for answering part $(b)$.

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  • $\begingroup$ @John.So answer to (a) and (b) is same because $P(A\cap{B}=\emptyset)=P(A\subset\overline{B})$ Am I right? $\endgroup$ – Dhamnekar Winod Jan 14 '16 at 14:24
  • $\begingroup$ @user261056 You can use part (a) to answer part (b). If A and B have no elements in common, what does that mean A is a subset of? $\endgroup$ – John Jan 15 '16 at 16:04

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