drawing two coins, what is the probability you get head on one coin, and tail on the other? 
there are three coins, probability that you get a head in the first coin is 0.5 (a fair coin), probability that you get head in the second coin is 0.75, and third coin is a two-headed coin. if you draw two coins randomly (without replacement), and you toss these coins, what is the probability you get exactly head on one coin, and tail on the other coin?

This exercise gives a tip: order doesn't matter, and therefore I have a combination.
I know that when I have the and operator, I have to do a product because it's an intersection, meaning the first must be true, and so the second.
I know it's without replacement too, and so it's an hypergeometric distribution. Hypergeometric distribution accepts three parameters, namely $(h, t, n)$, $h$ is head, $t$ is tails, and $n$ is  the number of coins I'm drawing.
hypergeometric distribution has a product in the numerator, and that product is define as $C(h, k) \cdot C(t, n-k)$, and at the denominator I have the sample space and it's defined as $C(h+t, n)$ (this notation means combinations).
I don't know how to use what I've written above to solve this assignment. what are values of $h, k$, and $t$? I'm having problems understanding the assignment.

Call the coins $A, B, C$ (resp. fair, biased, two-headed).
You have a $\frac13$ chance of picking any pair ($AB$, $AC$, or $BC$). If you pick $AB$ or $AC$, you have a $\frac12$ probability of getting one head and one tail (coin $A$ has to disagree with the other coin).
If you choose $BC$, you have a $\frac14$ probability of one head and one tail (coin $B$ must come up tails).
So $P($ 1 head, 1 tail $)=\frac23\cdot \frac12+\frac13\cdot\frac14=\frac{5}{12}$
