Conditional probability for the sum of drawn cards. Where am I going wrong? I have the following practice problem.

From a deck of five cards numbered 2, 4, 6, 8, and 10, a card is drawn at random and replaced. This is done three times. What is the probability that the card numbered 2 was drawn exactly two times, given that the sum of the numbers on the three draws is 12?

I keep ending up at the following:
$$3\cdot\left(\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}\right)=\frac{3}{125}$$
However, the provided answer is $\frac{3}{10}$.
My reasoning comes from doing a logic tree where you have $\frac{1}{5}$ probability of choosing 2, followed by $\frac{1}{5}$ probability of choosing 2, followed by $\frac{1}{5}$ probability of choosing 8 (in order to get a sum of 12). There are three permutations so I end up with my solution above.
Where am I going wrong?
 A: Hint: There are ten outcomes that give a sum of 12: (4,4,4), (2,2,8), (2,8,2), (8,2,2), (2,4,6), (4,2,6), (6,2,4), (2,6,4), (4,6,2), (6,4,2).
A: Let's do this formally, as a guard against error.  Let $A$ be the event that $2$ was drawn exactly twice, and let $B$ be the event that the sum was $12$.  We want $P(A|B)$, the probability that $A$ happened given that $B$ happened. We use the familiar formula
$$P(A|B)P(B)=P(A\cap B).$$
There are $5^3$ possibilities for the sequence of numbers drawn. All of these are equally likely. Of these $5^3$ possibilities, exactly $3$, namely $(2, 2,8)$, $(2,8,2)$, and $(8,2,2)$, fall in $A\cap B$. It follows that $P(A\cap B)=\dfrac{3}{5^3}$.
Computing $P(B)$ is a little more tedious. We need to count the number of outcomes with sum $12$. There are $3$ with two $2$s.  How many with one $2$? These are $(2,4,6)$ and the other $5$ permutations of it, for a total of $6$. And let's not forget $(4,4,4)$. The total is $10$, so $P(B)=\dfrac{10}{5^3}$.
It follows that
$$P(A|B)=\frac{\frac{3}{5^3}}{\frac{10}{5^3}}=\frac{3}{10}.$$  
Comment: What went wrong in your argument should be clear. You calculated correctly the probability that we get two $2$'s and an $8$.  In the notation above, you calculated $P(A\cap B)$.  And $A\cap B$ was pretty unlikely.  But being given the information that the sum is $12$ makes two $2$s and an $8$ much more likely. We are effectively confining attention to situations when $B$ happened. To get the correct conditional probability, we need to divide by $P(B)$.
A: This looks like a classic Bayes' rule problem. We want to find the probability of two 2s given that the three numbers sum to 12. We can apply Bayes' rule thusly:
$$
P(\text{two }2s| \text{sum}=12) = \frac{P(\text{sum}=12 | \text{two } 2s)*P(\text{two } 2s)}{P(\text{sum}=12)}
$$
It should be easier to find these probabilities than the original one directly (especially with the help of Byron Schmuland's answer). Note that the numerator is also $P(\text{two } 2s \bigwedge \text{sum}=12)$, which you have already found.
