probability of getting particular cards in a hand (cards) I was given the question in class:
"What is the probability of getting a hand with 1 heart, 2 diamonds, 2 clubs?" 
and 
"What is the probability of getting a hand with at least 3 queens"
for the first question I assumed that the cards were removed from the deck each time so i removed them from the total each time and multiplied each probability:
13/52 * 13/51 * 12/50 * 13/49 * 12/48
but i got an abnormally small answer to this and I'm not sure if i'm on the right track..
And I'm not 100% sure what do with the second question for "atleast" questions i generally find the compliment of the event and then use that.
Could I please get some help with these two questions?
 A: We assume that in each problem we are picking a $5$-card hand at random. 
In your first calculation, what you found is the probability  of getting $1$ heart, $2$ diamonds, and $2$ clubs in that order. 
The following approach will get you the right probability. There are $\binom{52}{5}$ $5$-card hands, all equally likely. 
There are $\binom{13}{1}\binom{13}{2}\binom{13}{2}$ hands with $1$ heart, $2$ diamonds, and $2$ clubs. For the heart can be chosen in $\binom{13}{1}$ ways. For each choice of heart, there are $\binom{13}{2}$ ways of choosing $2$ diamonds, and for each way of choosing the heart and diamonds, there are $\binom{13}{2}$ ways to choose the clubs. Thus the required probability is
$$\frac{\binom{13}{1}\binom{13}{2}\binom{13}{2}}{\binom{52}{5}}.$$

We find the number of $5$-card hands that have $3$ Queens, and the number that have $4$ Queens.
How many $3$-Queen hands are there? The Queens can be chosen in $\binom{4}{3}$ ways. For each way, we can choose the $2$ non-Queens to go with the Queens in $\binom{48}{2}$ ways. Thus there are $\binom{4}{3}\binom{48}{2}$ $3$-queen hands. 
How many $4$-Queen hands are there? A similar (but simpler) calculation shows this is $\binom{4}{4}\binom{48}{1}$, or more simply $48$. 
So the probability of $3$ or more Queens is $\dfrac{\binom{4}{3}\binom{48}{2}+\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$.
