Probability of at least one king and at least one ace when 5 cards are lifted from a deck? Here's what I've gathered so far: 
First I calculate the number of combinations of $5$ cards from the whole deck, which is $2 598 960.$
I'll use the complement, so I want to the combinations of everything but the aces and kings, so it's again combinations of $5$ from $44$ cards, which is $1086008$.
$$1-\frac{1086008}{2598960} = 0.582$$
That is incorrect however. What am I doing wrong? The complement for "at least one $X$ and at least one $Y$" should be "not $X$ or not $Y$", which is "everything but aces and kings". Is that even correct?
 A: We need to subtract from $C^{52}_{5}$ (the total number of ways of choosing $5$ cards from $52$) the number of ways of selecting no aces and the number of ways of selecting no kings, but we must add the number of ways of selecting no aces and no kings (else we'll have counted these twice). The number of ways of selecting neither aces nor kings is $C^{52-8}_{5}=C^{44}_{5}$. The number of ways of selecting no aces is $C^{52-4}_{5}=C^{48}_{5}$. This is equal to the number of ways of selecting no kings. So we have: $C^{52}_{5} - 2C^{48}_{5}+C^{44}_{5}$ ways of selecting $5$ cards from a deck such that we have at least one ace and at least one king. The probability of this is then $\dfrac{C^{52}_{5} - 2C^{48}_{5}+C^{44}_{5}}{C^{52}_{5}} \approx 10\%$.
A: Let $K,A$ denote the number of Kings and Aces drawn, respectively.
$$\begin{align*}
P(K>0 \text{ and } A>0) &= 1 - P(K=0 \text{ or } A=0) \\
&= 1 - P(K=0) - P(A=0) + P(K=0 \text{ and } A=0)
\end{align*}$$
The final term exists because we have subtracted that event twice, once in the term $P(K=0)$ and once in the term $P(A=0)$; now we have to add it back to account for our double-counting (this is the principle of inclusion-exclusion). The values in the last line are easy to calculate. $P(K=0) = P(A=0) = \binom{48}{5}/\binom{52}{5}$ and $P(K=0 \text{ and } A=0) = \binom{44}{5}/\binom{52}{5}$.
