Answer to Question #125662 in Statistics and Probability for ash

Question

A player would win a game if they draw exactly 2 Aces or exactly 2 Kings in a hand of 5 cards from a standard deck of 52 cards. Explain why it is necessary to use the Principle of Inclusion Exclusion or a Venn diagram to calculate the probability.

Answer 1

Solution:

We can calculate the probability that player draw exactly 2 Aces, the probability that player draw exactly 2 Kings and the probability that player draw exactly 2 Aces and exactly 2 Kings:

P(exactly 2 Aces in a hand of 5 cards)=m/n

m=binomial(4,2)*binomial(48,3)

binomial(4,2)=4!/2!/2!=6

binomial(48,3)=48!/3!/45!=17296

n=binomial(52,5)=2598960

P(exactly 2 Aces in a hand of 5 cards)=6*17296/2598960=2162/54145

P(exactly 2 Kings in a hand of 5 cards)=P(exactly 2 Aces in a hand of 5 cards)

P(exactly 2 Kings in a hand of 5 cards)=2162/54145

P(exactly 2 Aces and exactly 2 Kings in a hand of 5 cards)=m/n

m=binomial(4,2)*binomial(4,2)*binomial(44,1)

m=6*6*44=1584

n=binomial(52,5)=2598960

P(exactly 2 Aces and exactly 2 Kings in a hand of 5 cards)=1584/2598960=33/5145

And using the Principle of Inclusion Exclusion or a Venn diagram we can calculate the probability that player draw exactly 2 Aces or exactly 2 Kings:

P(exactly 2 Aces or exactly 2 Kings in a hand of 5 cards)=

=P(exactly 2 Aces in a hand of 5 cards)+P(exactly 2 Kings in a hand of 5 cards)-

-P(exactly 2 Aces and exactly 2 Kings in a hand of 5 cards),

P(exactly 2 Aces or exactly 2 Kings in a hand of 5 cards)=2162/54145+2162/54145-33/5145=

=613/7735

In this solution we use the binomial coefficient:

binomial(a,b)=a!/b!/(a-b)!