Answer to Question #284179 in Statistics and Probability for Fox

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

Solution:

$i)$

We apply combinations to find the 8 card hands can be dealt as follows,

number of ways =$\binom{n}{k}={n!\over(k!(n-k)!)}$

From the question, $n=52, k=8$

Therefore,

$\binom{52}{8}={52!\over(8!\times44!)}=752538194$

Thus, there are 752538194-8card hands that can be dealt.

$ii)$

To get exactly 2 aces, we need to choose 2 of the 4 aces and 6 of the other 48 cards. The number of ways to do that is

$\binom{4}{2}\times\binom{48}{6}={4!\over(2!*2!)}*{48!\over(6!*42!)}=6\times12271512=73629072$

Therefore, the probability is, ${73629072\over752538194}=0.0978(4dp)$

$iii)$

We first choose 7 cards out of a total of 13 cards in a suit. But we only want one out of 4 suits. Therefore, the number of ways for choosing 7 cards out of a total of 13 cards is $\binom{13}{7}={13!\over(7!\times6!)}=1716$ and the number of ways of choosing 1 out out 4 suits is, $\binom{4}{1}={4!\over(1!\times3!)}=4$

Total number of 1716*4=6864

Hence the probability is, ${6864\over752538194}=9.121132e^{-6}$

Therefore, the probability that a hand of 7 dealt randomly will have 7 cards of the same

suit is $=9.121132e^{-6}$