Computing card probabilities $1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

*

*All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.



*Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.


*Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.


*There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).


*All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.
Are my solutions correct? If not, where did I go wrong?

The first three and the last one are correct.
In the fourth one you’ve overcounted. Specifically, you’ve counted each set of $3$ that contains exactly two $12$s twice, once for each $12$, and each set of $3$ that includes exactly three $12$s three times, once for each $12$: any one of the $12$s can be the one that you singled out at the beginning, and you’re counting each of those possibilities separately.
It’s easier here to count the sets of $3$ that contain no $12$s or $13$s and subtract that from number of sets of $3$ that contain no $13$s. There are $\binom{44}3$ ways to choose $3$ cards that are neither $12$s nor $13$s, and there are altogether $\binom{48}3$ ways to choose $3$ cards that are not $13$s, so there are $\binom{48}3-\binom{44}3$ ways to choose $3$ cards that include at least one $12$ but no $13$s.
