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(1) Let $X$ be a set containing $n$ elements.The Subsets $A$ and $B$ of $X$ are choosen at Random , Then

the probability that $A\cup B = X$ , is

(2) Let $X$ be a set containing $n$ elements.The Subsets $A$ and $B$ of $X$ are choosen at Random , Then

the probability that $A\cap B = \phi$ , is

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up vote 2 down vote accepted

Consider the number $k$, where $1\le k\le n$. It is in $A$ with probability $\frac{1}{2}$, and in $B$ with probability $\frac{1}{2}$. So with probability $\frac{3}{4}$ it is in at least one of $A$ or $B$. This is clear, for the probability that $k$ is in neither $A$ nor $B$ is $\frac{1}{2}\cdot\frac{1}{2}$.

In order to have $A\cup B=X$, this has to happen $n$ times. The probability of that is $\left(\frac{3}{4}\right)^n$.

The same sort of argument works for intersection, and is left to you.

Remark: Alternately, for the second problem, we could recycle the result of the first problem. In general, let $Y^c$ be the complement of the set $Y$ (with respect to $X$). Then $A\cap B=\emptyset$ if and only if $A^c\cup B^c =X$. But $A^c$ and $B^c$ are randomly chosen sets, so again the probability is $\left(\frac{3}{4}\right)^n$.

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Thanks André Nicolas for giving me a solution. but i did not understand $A$ with a probability $\frac{1}{2}$ and $B$ with aprobability $\frac{1}{2}$ – juantheron Jan 31 '13 at 5:43
You are welcome. – André Nicolas Jan 31 '13 at 5:51
To André Nicolas would you like to explain me how can i get probability is $\frac{1}{2}$ for $A$ and $B$ Thanks – juantheron Jan 31 '13 at 15:39
The problem says subset $A$ is chosen "at random." That's not quite precise enough, but what is intended by at random is that each of the $2^n$ subsets of $\{1,2,\dots,n\}$ is equally likely. Let $1\le k\le n$. There are just as many subsets of $\{1,2,\dots,n\}$ that contain $k$ as there are that don't contain $k$, in each case $2^{n-1}$. So the probability $A$ contains $k$ is $\frac{2^{n-1}}{2^n}=\frac{1}{2}$. More intuitively, a random set $A$ is produced by looking at $1,2,\dots,n$ in turn, and flipping a fair coin to decide for each number whether it will be "in" or "out". – André Nicolas Jan 31 '13 at 15:59

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