Clearly, the total number of subsets possible is $2^5$
For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.
Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?
Even in complimentary method there seem to be a lot of subsets, but here is what I got
$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33)…..+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$
Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?
Edit: My reasoning
I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.
Best Answer
Note: $2^5$ is the count of whether each element is in a subset or not. However, you are selecting 2 such subsets. So we must count whether each element is included in each subset: which is $4^5$ outcomes in the space.
For the favored case, we wish a selection of 2 from the five elements to be in both, and each of the remaining 3 elements have but 3 options (in one, in the other, or in neither). So that is $\binom 52 3^3$ favored outcomes.
As all outcomes are unbiased, the probability is: $$\dbinom 5 2\cdot \dfrac{3^3}{4^5} =\dfrac{135}{512}$$