Probability of a drawing a specific suit and a specific color After drawing 5 cards from a standard 52 card deck, what is the probability that the hand will contain:


*

*1 diamond 

*1 spade

*Any other red card (diamond or heart)


My first approach was to use a hypergeometric distribution such as:
$$\frac{{13\choose 1}{13\choose 1}{26\choose 1}{???\choose 2}}{52\choose5}$$
However, this doesn't make sense because I have nothing left over to choose the remaining two.
What am I missing here?
 A: You wish to draw a five-card hand comprised of at least two red cards, at least one of which is a diamond, and at least one spade.
The ways to obtain this are to: Draw from 2 to 4 red cards that are not all hearts, and the remainder of the cards drawn are to be black but not all clubs.
$$\begin{align} 
&\text{Thus we count:}
\\ &\quad \frac{\sum\limits_{k=2}^4 \bigg({26\choose k}-{13\choose k}\bigg)\bigg({26\choose 5-k}-{13\choose 5-k}\bigg)}{52\choose 5}
\\[1ex] & = \frac{
 \bigg({26\choose 2}-{13\choose 2}\bigg)\bigg({26\choose 3}-{13\choose 3}\bigg)
+\bigg({26\choose 3}-{13\choose 3}\bigg)\bigg({26\choose 2}-{13\choose 2}\bigg)
+\bigg({26\choose 4}-{13\choose 4}\bigg){13\choose 1}
}{52\choose 5}
\\[1ex]
& = \frac{102167}{199920}
\\[1ex]
 & \approx 0.511
\end{align}$$

[Addendum]
Your method seems to be trying to count ways to: draw a $\diamondsuit$, draw a $\spadesuit$, draw another red card, draw any two cards.
This is fraught with over counting.   Consider just one example: How many ways would the hand $\rm 10\heartsuit, J\diamondsuit, Q\diamondsuit, K\diamondsuit, A\spadesuit$ be over counted by your method?
$$\begin{array}{llll}
\text{ draw a $\diamondsuit$} &\text{ draw a $\spadesuit$} & \text{ draw another red card} & \text{ draw any two cards }
\\ J\diamondsuit & A\spadesuit & 10\heartsuit & Q\diamondsuit, K\diamondsuit
\\ J\diamondsuit & A\spadesuit & Q\diamondsuit & K\diamondsuit, 10\heartsuit
\\ J\diamondsuit & A\spadesuit & K\diamondsuit & Q\diamondsuit, 10\heartsuit
\\ Q\diamondsuit & A\spadesuit & 10\heartsuit & J\diamondsuit, K\diamondsuit
\\ Q\diamondsuit & A\spadesuit & J\diamondsuit & K\diamondsuit, 10\heartsuit
\\ Q\diamondsuit & A\spadesuit & K\diamondsuit & J\diamondsuit, 10\heartsuit
\\ K\diamondsuit & A\spadesuit & 10\heartsuit & Q\diamondsuit, K\diamondsuit
\\ K\diamondsuit & A\spadesuit & Q\diamondsuit & J\diamondsuit, 10\heartsuit
\\ K\diamondsuit & A\spadesuit & J\diamondsuit & Q\diamondsuit, 10\heartsuit
\end{array}$$
