Question 203046
If you check out the various combinations of sums and products of the numbers from 1 to 6, I think you will find that the only way for the sum of two numbers to be greater than the product of those two is if one of the numbers is 1. In other words, when selecting from {2, 3, 4, 5, 6}, the sum of any two numbers selected is never greater than the product.<br>
So the problem is now: If two numbers are randomly chosen without replacement from {1, 2, 3, 4, 5, 6}, what is the probability that at least one of the numbers is 1? And the easiest way to find this is to find the probability that <b>no</b> 1's are selected and then subtracting that from 1:
P(at least 1 one) = 1 - P(no ones)
P(no ones) = (5/6)*(5/6) = 25/36
P(at least 1 one) = 1 - 25/36 = 11/36
P(sum greater than product) = P(at least 1 one) = 11/36