# Five cards are dealt from a standard shuffled deck. What is the probability that:

All are of the same suit?
This would be $\frac{{13 \choose 5}}{{52 \choose 5}}$ would it not?

There is at least one card from each of the four suits?
I cannot figure out how to begin this one.

c) There is no more than one card from any of the four suits?
I know this is 0 because of the pigeonhole principle.

• The first card can be of any suit. That leaves 51 cards, 12 of which are from the same suit as that first card. The probability that the second card is the same suit is 12/51. Given that, there are now 50 cards in the deck, 11 from the same suit as the previous two. Then there are 49 cards. 10 from that suit, and finally, 9 cards out of 48. The probability of 5 cards of the same suit is (12/51)(11/50)(10/49)(9/48). – user247327 Nov 4 '16 at 0:39
• At least one from each of the four suits: The first can be anything. Then there are 51 cards left, 39 from a different suit as the first. Then there are 50 cards left, 26 from a different suit as the first two. Then there are 49 cards left, 13 from a different suit as the first three. The last card can be anything. The probability of "at least one card from any of the four suits" is (39/51)(26/50)(13/49) – user247327 Nov 4 '16 at 0:50
• Yes, since there are only 4 suits, it is impossible to have "no more than once card from each suit" In 5 cards. – user247327 Nov 4 '16 at 0:51

(1) Almost. You have to choose the suit> $$\dfrac{\dbinom 41\dbinom {13}5}{\dbinom {52}5}$$

(2) What is the probability of selecting two cards from one suit, and one card from each of the three remaining suits? (As above, select the suit for two cards)

$$\dfrac{\binom{4}{1}\binom{13}{2}\cdot\binom{13}{1}^3}{\binom{52}{5}}$$

Alternatively: Use complements, and PIE.

Use the Principle of Inclusion and Exclusion to find is the probability of not having all four suits;

$$1-\dfrac{\binom 4 3\binom{39}{5}-\binom 4 2\binom{26}{5}+\binom4 1\binom{13}{5}}{\binom{52}5}$$

(3) Yes; indeed!   It is rather difficult to select 5 cards of different suits from 4 suits.