Answer to Question #278693 in Statistics and Probability for THEASAMOAH

Question

You have a deck of 52 playing cards

(i) How many different 8 card hands can be dealt?

(ii) What is the probability that a hand of 8 dealt randomly contains (exactly) 2 aces?

(iii) What is the probability that a hand of 7 dealt randomly will have 7 cards of the same

suit?

Answer 1

(i)

$$\dbinom{52}{8}=\dfrac{52!}{8!(52-8)!}=752538150$$

(ii)

There are 4 aces in a deck of 52 playing cards

$$P(2Aces\ and\ 6\ nonAces)=\dfrac{\dbinom{4}{2}\dbinom{52-4}{8-2}}{\dbinom{52}{8}}$$

$$=\dfrac{6(12271512)}{752538150}\approx0.09784$$

(iii)

There are 4 suits in a deck of 52 playing cards.

There are 13 cards in each suit.

$$P(7\ the\ same\ suit)=\dfrac{\dbinom{4}{1}\dbinom{13}{7}}{\dbinom{52}{7}}$$

$$=\dfrac{4(1716)}{133784560}\approx0.0000513$$