Card game probability with simultaneous drawing From a usual deck of 52 cards we draw 10 cards simultaneously (thus no replacement). What is the probability to have 5 diamonds and a queen?
My attempt: the cardinality of all 10 card possibilities is: $52 \choose 10$. The probability of picking a queen is $4/52$ and the probability of picking 5 diamonds is $13 \choose 5$, finally $36 \choose 4$ ways of picking the last 4 cards. Overall we have then: 
$$P=\frac{4/52 * {13 \choose 5} * {36 \choose 4}}{52 \choose 10} = \frac{891}{2417116}$$
Is this correct? Have I mis-counted something due to the fact that among the 5 diamonds the queen of diamond could have already occured (and thus satisfied the queen condition as well)?
I'm having some difficulties as to how translate such questions in terms of union or intersection of simpler events, any help would be much appreciated.
 A: If I call $D$ the random variable to have some number of diamonds in the hand (ten cards) and $Q$ the random variable to have some number of queen in the hand the probability that I want is
$$\Pr[D=5\cap Q=1]$$
We have two different kind of hands for this situation and $D$ and $Q$ are not independent
$$\Pr[D=5\cap Q=1]=\begin{cases}\binom{12}{5}\binom{3}{1}\binom{52-16}{4}\binom{52}{10}^{-1},\ &\text{ if the queen is not diamond}\\\binom{12}{4}\binom{1}{1}\binom{52-16}{5}\binom{52}{10}^{-1},\ &\text{ if the queen is diamond}\end{cases}$$
Add the two cases to have the probability that you want
A: The likelihood of getting exactly k diamonds in ten cards is ${{13 \choose k}*{39 \choose 10-k}\over {52 \choose 10}}$.
The odds of getting at least one queen are not independent of the number of diamonds. (For example, if you have 13 diamonds you are sure to have a queen.) Rather, if you have k diamonds, the likelihood that none of them are the queen is ${(13-k)\over 13}$, while the likelihood of the non-diamonds all being non-queen is ${36!/(26+k)!\over 39!/(29+k)!}$.  Therefore, the likelihood of having at least one queen given k diamonds is 1 - ${(13-k)\over 13}*{36!/(26+k)!\over 39!/(29+k)!}$
That being said, the likelihood of getting at least 5 diamonds and at least one queen is $$\sum_{k=5}^{13} {{13 \choose k}*{39 \choose 10-k}\over {52 \choose 10}}*\left(1 - {(13-k)\over 13}*{36!/(26+k)!\over 39!/(29+k)!}\right)$$
