Probability of Sets

Question

I need some help on this one:

We have sets $X$ and $Y$ chosen independently and uniformly at random from among all subsets of $\{1,2,\ldots,100\}$. Determine the probability that $X$ is a subset of $Y$.

@JLee : If they were, then it would seem silly to phrase the question in the way in which it was phrased here. — Michael Hardy, Jan 08 '15 at 18:14
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score 26 · Answer 1 · answered Jan 08 '15 at 18:08

For every element $a$ of initial set one of four options is true:

$\quad a \in X,\quad a \in Y$
$\quad a \not\in X,\quad a \in Y$
$\quad a \in X,\quad a \not\in Y$
$\quad a \not\in X,\quad a \not\in Y$

The probability of each option is $1/4$ (due to independence and uniformity).
$X\subseteq Y$ if for all $a \in \{1,\ldots,100\}$ options number $1$, $2$ or $4$ are true. For every element probability is $\dfrac{3}{4}$ and for whole set it is $\left(\dfrac{3}{4}\right)^{100}$.

this is better. +1, I'll leave mine to see how this is better. — Asinomás, Jan 08 '15 at 18:09

Asinomás · Answer 2 · 2015-01-08T18:18:17.900

5

how many pairs $(X,Y)$ satisfy $X\subseteq Y$? classify according to the size of $Y$.

So we get $\sum\limits_{k=0}^n\binom{n}{k}2^k$.

Since there are $2^n\times 2^n$ possible pairs $(X,Y)$ you want $$\frac{\sum\limits_{k=0}^n\binom{n}{k}2^k}{2^{2n}}$$

Now notice that $\sum\limits_{k=0}^n\binom{n}{k}2^k$ is equal to $3^n$ because it is the number of ways to color the integers from $1$ to $n$ with colors white grey and black. To see this we could say we are indexing on the number of elements that are grey or black (So first pick the subset of numbers that are grey or black) And once that subset has been selected choose the subsets of that subset that is going to be black. Hence uniquely determining the coloring.

edited Jan 08 '15 at 18:18

answered Jan 08 '15 at 17:52

Asinomás

105,651

4

The numerator of this expression can be seen to be $(2+1)^n=3^n$. – paw88789 Jan 08 '15 at 18:12
oh yeah, I think I saw it in a book, I'll try to prove it combinatorially to keep with the counting theme in the answer – Asinomás Jan 08 '15 at 18:14
1

So $3^n/2^{2n}$, which is $(3/4)^n$. ${}\qquad{}$ – Michael Hardy Jan 08 '15 at 18:16
I think I found a non too shabby proof that the numerator is $3^n$ – Asinomás Jan 08 '15 at 18:18

score 0 · Answer 3 · answered Jan 09 '15 at 13:36

Let $D_n = \{i| i \in \Bbb N \land i \le n\}$.

For an empty set $D_0$ all its subsets $X,Y \subseteq D_0$ are empty: $X = Y = \emptyset$, so $P_0(X\subseteq Y) = 1.$

Now let's consider $D_n$ for $n\in \Bbb N$. Then any subset $X\subseteq D_n$ either is a subset of $D_{n-1}$ or it contains the number $n$, both with probability equal $\tfrac 12$. For subsets $X,Y$ there are four equally possible cases of $n$ inclusion, which results in $$P_n=\tfrac 14P_{n-1} + 0 + \tfrac 14P_{n-1} + \tfrac 14P_{n-1} = \tfrac 34P_{n-1} $$ Since $P_0 = 1 = \left(\tfrac 34\right)^0$, the above makes a general answer $$P_n=\left(\tfrac 34\right)^n$$

Probability of Sets

3 Answers3