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This is a 5 card poker hand. For this problem is it okay to take the complement and find the probability that you get no face cards and no diamonds? Thus since we have 13 diamonds and 8 non-diamond face cards we get:

${21 \choose 5} / {52 \choose 5}$

and then do 1 - ${21 \choose 5} / {52 \choose 5}$ to get the probability of at least one face card and at least one diamond?

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  • $\begingroup$ The complement of $A$ and $B$ is not $A$ or not $B$, which becomes what @Kiran wrote. $\endgroup$ – Tim Thayer Feb 4 '17 at 17:37
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Calculate it using inclusion-exclusion principle, as

P(at least one face card and at least one diamond) = 1 - P(no face card) - P(no diamond) + P(no face card and no diamond card)

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