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The problem in full is:

If 10 voters are for A, 8 voters are for B, and 6 voters are for C, what is the probability that a random selection of 6 (no two can be the same) voters will yield 2 voters for each candidate?

and a follow up is

Instead, suppose you call 6 numbers chosen randomly (same number can be chosen) from the pool of 24 voters--now what is the probability that the calling will yield 2 voters for each candidate?

my work so far is as follows:

for the first:

(C(10,2)*C(8,2)*C(6,2) / 3!) / (24*23*22*21*20*19)

and for the second:

(C(10,2)*C(8,2)*C(6,2) / 3!) / (24^6)

my thought process is that you must get 2 from each group, you must divide by 3! to eliminate combinations including the same people in different order, and the denominator is the total number of ways to choose/call 6 people in each different case.

Thanks for any help in advance!

1 Answers 1