The set will be

S = {2, 3, 5, 7}

Two numbers can be chosen from the set of four numbers in

4C2 = 6 ways.

Chosen Numbers ----- Product ----- Numbers not chosen ----- Product

2, 3 -------------------- 6 --------------- 5, 7 -------------------- 35

2, 5 -------------------- 10 --------------- 3, 7 -------------------- 21

2, 7 -------------------- 14 --------------- 3, 5 -------------------- 15

3, 5 -------------------- 15 --------------- 2, 7 -------------------- 14

3, 7 -------------------- 21 --------------- 2, 5 -------------------- 10

5, 7 -------------------- 35 --------------- 2, 3 -------------------- 6

Thus, from the table above, it is obvious that there are just three possible combinations in which the product of the numbers chosen is greater than the product of the numbers not chosen.

Thus the required probability = 3/6 = 1/2

Hope this is helpful.