Saw this somewhere

Each element of $X$ has an equal chance of either being in or out of a given subset of $X$ . So the probability that an element of $X$ is in neither of $A$ or $B$ is $(\frac{1}{2})^{2} = \frac{1}{4}$ , and thus the probability that this element is in at least one of $A$ or $B$ is $1 - \frac{1}{4} = \frac{3}{4}$ . As this is the case for each of the $n$ elements of $X$ , the probability that every element of $X$ is in at least one of $A$ or $B$ is $(\frac{3}{4})^{n}$ , and so $k = \frac{3}{4} = \boxed{0.75}$ .

First, we'll pick a subset A with k elements. There are $2^k$ to pick from. The probability that we'd pick another subset B that is one of these substets is given by:

$\frac { { 2 }^{ k } }{ { 2 }^{ n } } ={ 2 }^{ k-n }$

But that's only for one subset with k elements. The probability of picking such subset is:

${ 2 }^{ -n }\left( \begin{matrix} n \\ k \end{matrix} \right)$

Now, sum over all possible values of k: $\sum _{ k=0 }^{ k=n }{ { 2 }^{ k-n } } { 2 }^{ -n }\left( \begin{matrix} n \\ k \end{matrix} \right)$

Using: ${ (x+y) }^{ n }=\sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } { x }^{ n-k }{ y }^{ k }$

We get that: $\sum _{ k=0 }^{ k=n }{ { 2 }^{ k-n } } { 2 }^{ -n }\left( \begin{matrix} n \\ k \end{matrix} \right) ={ \left( \frac { 3 }{ 4 } \right) }^{ n }$

k = 3/4 = 0.75