# 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?

• Hint: Add the probabilities that a specified suit is missing, then subtract the probabilities that two specified suits are missing. – lulu Jun 6 '18 at 12:25
• Welcome to MSE. Please consider using MathJax for mathematical formulae and terms: math.meta.stackexchange.com/questions/5020/… – Tony Hellmuth Jun 6 '18 at 12:27
• The spades are missing???? – Mohammad Zuhair Khan Jun 6 '18 at 13:32
• @lulu ok, so I am using the Inclusion-Exclusion rule? So the probability is: P(heart missing U club missing U diamond missing) - P(heart&club missing U heart&diamond missing U club&diamond missing)? Can you explain why you cannot just subtract 1- the complement? – Joanne Jun 6 '18 at 13:42
• You can use the complement but there are many more separate cases that go into the complement than there are doing it by inclusion-exclusion. The formula you wrote, even if you "fix" it with a (8C1), still over-counts the complement since e.g., in a hand with 2 cards from each of the three suits, which 3 are the suit-specific chosen cards vs. the (25C3) cards? – Ned Jun 6 '18 at 15:52

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.