Intersection of two Combinations for Card Hand Probability

The question provided is:

A deck of 52 cards has 13 spades, 13 hearts, 13 diamonds, and 13 clubs. Suppose that you are dealt a hand that consists of 5 cards. What is the probability that the hand has exactly 2 spades or exactly 2 hearts?

What I have so far is:

$$P(A) = \frac{|A|}{|S|}$$ $$P(A) = P((\text{2 spades in hand}) \cup (\text{2 hearts in hand}))$$ $$\text{inclusion and exclusion principle}$$ $$P(A) = P(\text{2 spades in hand}) + P(\text{2 hearts in hand}) - (P(\text{2 hearts in hand}) \cap P(\text{2 spades in hand}))$$

$$P(A) = \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} + \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }} - ( \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} \cap \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }})$$

Now I am not sure how to intersect the the two hands. But here is what I got:

$$P(A) = \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} + \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }} - \frac{{26\choose 3}_{\text{diamonds and clubs}}}{{ 52 \choose 5 }}$$

Which I then solve and get:

$$P(A) = 0.547559$$

I am not sure I intersected the two hands correctly so any advice to that regard would be much appreciated.

Thanks.

• In a great deal of the beginning you use "spades" twice and do not use "hearts" – JMoravitz Feb 13 '18 at 21:34

It is $$P(2\text{ spades and }2\text{ hearts})=\frac{\binom{13}{2}\binom{13}{2}\binom{26}{1}}{\binom{52}{5}}$$ otherwise it seems correct.