Set S consists of all prime numbers less than 10.

There are four prime numbers less than 10 — 2, 3, 5, and 7. We are selecting 2 numbers at random — this gives us 6 possible pairs:

7 and 5
7 and 3
7 and 2
5 and 3
5 and 2
3 and 2

So how many pairs will yield a product greater than the product of the remaining numbers? Let's start with the pairs with the largest numbers and then move to the pairs with the smallest numbers.

7 * 5 > 3 * 2 YES
7 * 3 > 5 * 2 YES
7 * 2 < 5 * 3 NO

5 * 3 > 7 * 2 YES
5 * 2 > 7 * 3 NO

3 * 2 < 7 * 5 NO

So 3 out of 6 pairs yield a greater product. So the probability that we will pick one of those pairs is 3/6 = 1/2.

Note: we could have used a combination to solve for the total number of possible pairs. We have 4 pick 2:

$$\frac{4!}{\left(4-2\right)!2!}=\frac{4\cdot3}{2}=6$$

So there are 6 possible pairs. However, since there were so few prime numbers under 10, yielding so few pairs, AND since we needed to know what the pairs were to evaluate their products, writing out the pairs is not a bad call on this particular question. In general, though, using permutation and combination equations will save you time.