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
(C) 17/28
(D) 3/4
(E) 301/336

This is on of the questions from "Challenge Problems" section.
My problem with the solution is that for calculating the "probability that the produc..." the permutations (6) are considered, but not so for the summation part. why? If the product can be obtained with numbers 2, 3 and 5 in 6 different ways, so can the sum resulting in an odd number, right?

There are 8 primes in set S: 2, 3, 5, 7, 9, 11, 13, 17, and 19.

1) Probability that the product of randomly chosen three numbers is a number less than 31
As can be seen, only 2, 3, and 5 can be combined to make the product less than 31. The number of ways is 3*2*1 = 6.

2) Probability that the sum of randomly chosen three numbers is odd.
I guess in this case you misunderstand the question stem when assuming that the three numbers whose product is less than 31 (2, 3, and 5) are the same as the three numbers whose sum is odd, but they are not.
As can be seen, all items in set S are odd, except 2. The sum of any three odd numbers is odd; therefore, the three odd numbers which can be combined to make the sum odd is chosen from the 7 remaining numbers of the set. The number of ways is: 7*6*5.

Hope this helps.

To start with

Set S is {2,3,5,7,11,13,17,19}


The probability that the the product of the three numbers is a number less than 31 only occurs when the three smallest numbers are chosen (i.e. 2 x 3 x 5 = 30); any other combination of three numbers from Set S will yield a product greater than 31.

The probability that all three numbers are 2, 3, 5 is
(3/8) x (2/7) x (1/6) = (1/56)

Explanation for this:
(3/8) represents the probability 2, 3, or 5 is chosen on the first draw.

(2/7) represents the probability that one of the two remaining numbers is chosen. Remember, each number can be chosen once, so there is no replacement.

(1/6) represents the remaining number that must be chosen to have a product less than 30.

So, the probability that three numbers have a product less than 31 is 1/56


The probability that the sum of three numbers from Set S is odd: We have 7 odd numbers and 1 even in number in our set. So, we have two possibilities:

odd + odd + odd = odd
odd + odd + even = even

Thus, the only way to have the sum of the numbers be odd is if we pick only odd numbers.

The probability of picking only odd numbers is.
(7/8) x (6/7) x (5/6) = (5/8)

Difference between the two probabilities:
= (5/8) - (1/56)
= (35/56) - (1/56)
= (34/56)
= (17/28)

So, I am pretty sure the answer is C.

_________________
Rajiv Bhatia
www.gmathints.com

good analysis. ideally an efficient solution would take the same approach with both of these calculations, but either calculation can be performed either taking order into consideration or not. i recommend you solve it both ways in practice so you can get experience with both methods..

_________________
Tim Sanders
Manhattan GMAT Instructor

= 7C3/8C3 - 1/8C3
= 17/28

7C3, since except the number 2 all additions gives an odd number... 1 because only 2*3*5<30

I understand Kapur's question to be, "If the numerators in the separate probabilities are 2, 3, and 5, how can the numerator in the product of probabilities be odd?" Of course you might just as well have asked "How can it NOT be a multiple of 3?", "How can it NOT be a multiple of 5?"

There are two answers implicit in the various responses:

1) The 2, 3, and 5 represent the probabilities of the UNFAVORABLE outcomes, this is a 1-x approach.

2) The numerator in the simplified fraction will not have all the factors of the numerator in the unsimplified fraction.



Correct me if I'm wrong.