Deck of cards probability with exclusion What is the probability that a hand of five cards has exactly one club or exactly one heart
So my logic was to select all possibilities with one club + all possibilities with one heart - possibilities with one heart and one club, divided by all possible hands and ended up with
$\frac{(2*{13 \choose 1}{39\choose 4})-{13 \choose 1}{13 \choose 1}{26 \choose 4}}{52 \choose 5}$
edit: this seems really small so it is
${(2*{13 \choose 1}{39\choose 4})-{13 \choose 1}{13 \choose 1}{26 \choose 4}}$ divided by ${52 \choose 5}$
Is this correct? and if so is it possible to simplify this?
 A: Let $A$ be the event that I have exactly one club in my hand and $B$ be the event that I have exactly one heart in my hand. We see by inclusion-exclusion that
$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$
As you stated before, we see that if I have the event that I have exactly one club in my hand, there are $13$ ways of choosing that club, and for the other $4$ cards, there are $\binom{52-13}{4} = \binom{39}{4}$ ways of getting the rest of these cards. Given that our sample space is of size $\binom{52}{5}$, you are correct in stating that
$$P(A) = \frac{13\binom{39}{4}}{\binom{52}{5}}.$$
In an analogous manner,
$$P(B) = \frac{13\binom{39}{4}}{\binom{52}{5}}.$$
If I have exactly one club and exactly one heart in my hand, there are $13$ ways of choosing that club and $13$ ways of choosing that heart. given that there are $\binom{52-2 \times 13}{3} = \binom{26}{3}$ ways of choosing the rest of the cards,
$$P(A \cap B) = \frac{(13)^2 \binom{26}{3}}{\binom{52}{5}}.$$
Hence,
$$P(A \cup B) = \frac{2 \times 13\binom{39}{4} - (13)^2\binom{26}{3}}{\binom{52}{5}}.$$
Technically, there is a way to factor $26$ out of both the numerator and denominator, but this gives you a fraction that is not necessarily more readable then the unsimplified solution.
A: Yes, that looks about right, with correction for a possible typographic errro. $$\begin{align} & \mathsf P(\{\heartsuit=1\}\cup\{\clubsuit=1\})\\[1ex] ~=~& \mathsf P(\{\heartsuit=1\})+\mathsf P(\{\clubsuit=1\})-\mathsf P(\{\heartsuit=1\}\cap\{\clubsuit=1\}) 
\\[1ex] ~=~ & \dfrac{2\times\binom{13}{1}\binom{39}{4}-\binom{13}{1}^2\binom{26}{\color{red}3}}{\binom{52}{5}}
\\[1ex] =~& \dfrac{65351}{99960}
\end{align}$$
