What is the probability that at least one suit is missing from the selection? A deck of cards contains 10 hearts, 10 clubs and 8 diamonds. If 6 cards are selected at random, without replacement, what is the probability that at least one suit is missing from the selection?
I know that we must use the complement of the probability listed above, which is the probability that no suits are missing/all suits chosen. I calculated...
1- (10C1)(10C1)(25C3)/(28C6)... but this isn't correct. Any suggestions?
 A: If you are adamant that you must compute using the complement then you can proceed as follows.
Solution. Before proceeding forward lets us first defined our events. Let $\mathcal{N}$ denote the event that at least one suit is missing from the selection. In addition define the event $H_\alpha$  such that of the $6$ cards selected $\alpha$ were hearts   where $\alpha\in\{1,2,3,4\}$ and observe that
$$\mathbf{P}(\mathcal{N}^c) = \sum_{\alpha\text{ }=\text{ }1}^{4}\mathbf{P}(\mathcal{N}^c\cap H_\alpha)$$
Now observe that if only $1$ of the $6$ cards selected is a heart then all possibilities for clubs and diamonds are as follows.
$$X_1 = \{(1,4),(2,3),(3,2),(4,1)\}$$
where the number in the first slot represents the number of clubs chosen and that in the second slot represents the number of diamonds chosen.
Similary for $\alpha = 2,3,4$, $X_2 = \{(1,3),(2,2),(3,1)\}$, $X_3 = \{(1,2),(2,1)\}$ and $X_4 = \{(1,1)\}$
Consequently the probability of interest is as follows
$$\mathbf{P}(\mathcal{N}) = 1 - \sum_{\alpha\text{ }=\text{ }1}^{4}\sum_{(\beta,\gamma)\in X_\alpha}\frac{\binom{10}{\alpha}\binom{10}{\beta}\binom{8}{\gamma}}{\binom{28}{6}}$$

I am sure you can handle the rest.
