Bunuel wrote:
A lottery game consists of the host removing one ball at a time from an opaque jar. Each ball has one of the digits (0 – 9) written on it, and no two have the same number on them. If the host removes three balls without replacement, what is the probability that the sum of the numbers written on the balls equals 24?
(A) 1/3
(B) 3/10
(C) 1/720
(D) 1/120
(E) 1/40
As written, there's no way to answer the question. It doesn't tell us how many balls we have. We might, for example, have just three balls with the digits 7, 8 and 9 on them -- that situation is completely consistent with everything the question says. Then if we pick 3 balls, the sum will certainly be 24, and the answer is 1.
They presumably mean there are ten balls in the jar with the digits 0-9 on them. Then there are (10)(9)(8)/3! = 120 possible sets of 3 balls we could make, and only 1 set that gives us the sum we want (we must pick 7, 8, and 9 to get a sum as big as 24), so the answer is 1/120.
Another way to look at it: we know we must pick 7, 8 and 9 in some order. There is a 3/10 chance the first selection is one of these, then a 2/9 chance the second will be one of the two remaining numbers we need, and a 1/8 chance the third will be the last remaining number we need, so the answer is (3/10)(2/9)(1/8) = 1/120