If set S is the set of all prime integers between 0 and 20 then:
S = {2, 3, 5, 7, 11, 13, 17, 19}
Let's start by finding the probability that the product of the three numbers chosenis a number less than 31. To keep the product less than 31, the three numbers must be 2, 3 and 5. So, what is the probability that the three numbers chosen will be some combination of 2, 3, and 5?
Here's the list all possible combinations of 2, 3, and 5:
case A: 2, 3, 5
case B: 2, 5, 3
case C: 3, 2, 5
case D: 3, 5, 2
case E: 5, 2, 3
case F: 5, 3, 2
This makes it easy to see that when 2 is chosen first, there are two possible combinations. The same is true when 3 and 5 are chosen first. The probability of drawing a 2, AND a 3, AND a 5 in case A is calculated as follows (remember, when calculating probabilities, AND means multiply):
case A: (1/8) x (1/7) x (1/6) = 1/336
The same holds for the rest of the cases.
case B: (1/8) x (1/7) x (1/6) = 1/336
case C: (1/8) x (1/7) x (1/6) = 1/336
case D: (1/8) x (1/7) x (1/6) = 1/336
case E: (1/8) x (1/7) x (1/6) = 1/336
case F: (1/8) x (1/7) x (1/6) = 1/336
So, a 2, 3, and 5 could be chosen according to case A, OR case B, OR, case C, etc. The total probability of getting a 2, 3, and 5, in any order, can be calculated as follows (remember, when calculating probabilities, OR means add):
(1/336) + (1/336) + (1/336) + (1/336) + (1/336) + (1/336) = 6/336
Now, let's calculate the probability that the sum of the three numbers is odd. In order to get an odd sum in this case, 2 must NOT be one of the numbers chosen. Using the rules of odds and evens, we can see that having a 2 would give the following scenario:
even + odd + odd = even
So, what is the probability that the three numbers chosen are all odd? We would need an odd AND another odd, AND another odd:
(7/8) x (6/7) x (5/6) = 210/336
The positive difference between the two probabilities is:
(210/336) - (6/336) = (204/336) = 17/28
The correct answer is C.
