tnkippen wrote:I have been trying to use combinations more to solve probability questions, but I am finding a discrepancy between the combinations approach and the traditional approach on the problem below.

"There are 6 balls: A, B, C, D, E, F. What is the probability that a package containing two different types of balls contains E?"

Using combinations, I calculated: (5C5)(5C1)/(6C2) = 5/15 = 1/3

But using P(A or B) = P(A) + P(B) - P(A and B), I got:
P(A) = 1/6, P(B) = 1/5, P(A and B) = 0, so:
(1/6) + (1/5) = 11/30

What am I doing wrong above?

The problem is highlighted above (in green)

P(E is 1 of the 2 balls) = P(E selected first OR E selected second)
= P(E selected first) + P(E selected second) - P(E selected first and second)
= P(E selected first) + P(E selected second) - 0 (since E cannot be selected 1st and 2nd)

P(E selected first) = P(E selected 1st AND some other letter selected 2nd)
= P(E selected 1st) x P(some other letter selected 2nd)
= (1/6)(5/5)
= 1/6

P(E selected second) = P(some other letter selected 1st AND E selected 2nd)
= P(some other letter selected 1st) x P(E selected 2nd)
= (5/6)(1/5)
= 1/6

So, P(E is 1 of the 2 balls) = 1/6 + 1/6 = 1/3

Cheers,
Brent