TITLE: Second marble is of same color
QUESTION [5 upvotes]: A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. If replacement is not allowed, what is the probability that the second marble drawn will be red?
I disagree with the given answer $\frac29$. Can someone please point why my solution is wrong? 
P(Both Red) + P(Second is Red): 
P(Both Red) $=\frac{\binom22}{\binom92}$
P(Second is Red) $=\frac{\binom71\binom21}{\binom92}$
This comes out to be $\frac5{12}$.

REPLY [3 votes]: What other answers don't point out is why your reasoning is wrong in the first place.
The problem is that $\binom92$ is for an unordered set of 2 elements taken from 9-element set. But in your case you have to have an ordered set since only the second element of the choice is relevant and so the order is relevant.
The correct number of possibilities in such case for m elements chosen from n-element set is $\frac{n!}{(n-m)!}$ (and respectively for your needs:)

$\frac{9!}{(9-2)!} = 9*8$ for all possibilities
$\frac{2!}{(2-2)!} = 2$ for two reds
$\frac{7!}{(7-1)!} = 7$ for first non-red
$\frac{2!}{(2-1)!} = 2$ for second red when first wasn't red.

If you put that into your equation you'll get exactly the same result as provided by others, ending up with the final $\frac29$ result.