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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 $$A \subseteq B \subseteq C$$.

[MY ATTEMPT: I tried breaking the set into 4 parts, based on if it is in A,B, or C... not sure if this logic works or how to continue. Open to new ideas that don't use this line of thinking at all]

Mar 27, 2022
edited by Guest  Mar 27, 2022

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#2
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I've validated it... it is incorrect. Sorry. Any ideas?

Guest Mar 29, 2022
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WEll I have very little idea but this is where my play took me.

How many subsets of S are there altogether.    This is the number of ways that A can be chosen

$$\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}$$

subsets b and C can be shosen the same way so

number of ways subsets A,B and C can be chosen is       $$\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3$$

But what if  A is a subset of B and B is a subset of C ...

Ways to chose C is    $$\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}$$

then

ways to chose B is     $$\displaystyle \sum_{m=0}^k\;\;\binom{k}{m} \qquad m\le k$$

ways to choose A is    $$\displaystyle \sum_{t=0}^m\;\;\binom{m}{t}\qquad t\le m$$

Put thse toghether and I get the number of ways   $$A \subseteq B \subseteq C$$

is    $$\displaystyle \left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\$$      that is probably not displayed in the right order. So prob would be

$$\frac\left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\ }{\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3}$$

I don't know how to display that properly, the order and the brackets are most likely wrong.......

Mar 30, 2022