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How many five-card hands dealt from a standard deck of $52$ playing cards are all of the same suit? If a random hand is dealt, what is the probability that it will have this property?
Would the probability be: $$\frac{\dbinom{13}{5}*\dbinom{4}{1}}{\dbinom{52}{5}}$$
It would be $$\frac{\binom41_{\text{color}} \cdot \binom{13}5_{\text{cards of this color}} \cdot \binom{52-13}0_{\text{other cards}}}{\binom{52}{5}_{\text{total}}} = \frac{\binom41 \cdot \binom{13}5}{\binom{52}5} = \frac{33}{16660}$$ So you are correct. See here for some more on the number.
The first card can be any suit. For the second card there are 12 left of that suit out of 51 cards. For the third card there are 11 left of that suit out of 50 cards. For the fourth card there are 10 left of that suit out of 49 cards. For the fifth card there are 9 left of that suit out of 48 cards. So $ \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} = .001981 $ (same answer as another solution).
