jagdeep wrote:Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S
and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
[spoiler](C) 17/28[/spoiler](D) 3/4
(E) 301/336

The set of all Prime numbers under 20 is {2,3,5,7,11,13,17,19}

The total number of ways to choose 3 num out of it is 8C3=56

Condition 1: product of these numbers selected is less than 31 only one combination 2,3,5 satisfies this condition so the probability is 1/56

Condition 2: Since all of them are odd except 2 the sum is odd for all combinations except which contain 2
there are 7C3 =35 combination which doesnt contain 2 so the probability = 35/56


Now difference = 35/56-1/56 = 34/56 = 17/28

jagdeep wrote:Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S
and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?

(A) 1/336
(B) 1/2
[spoiler](C) 17/28[/spoiler](D) 3/4
(E) 301/336

There are 8 distinct primes in the given range, namely 2, 3, 5, 7, 11, 13, 17, and 19. Total number of ways of choosing 3 primes out of it, is 8*7*6.

2*3*5 is the only combination that ensures that the product of these three numbers is a number less than 31, which can be had in 3! = 6 ways. Hence, the probability that the product of these three numbers is a number less than 31, is 6/ (8*7*6) = 1/56.

All in set S, EXCEPT 2, are odd; sum of any three odds is an odd. Hence, the three primes from the set S, which ensure that the sum of these three numbers is odd, must be selected from the remaining 7 (MUST exclude 2 for a very good reason). This can be done in 7*6*5 number of ways, and hence the probability that the sum of these three numbers is odd, is (7*6*5)/ (8*7*6) = 5/8.

Since, 5/8 > 1/56, hence, the positive difference between the probabilities = (5/8) - (1/56) = [spoiler]17/28[/spoiler]. [spoiler]C[/spoiler]

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.