Question

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
(A) 833833

(B) 62656265

(C) 17331733

(D) 103165103165

(E) 2533

Answer 1

Answer:

$\frac{17}{33}$

Explanation:

The probability of finding at least on pair is equal to 1 minus the probability of finding no pair. If x represents the number of pairs obtained in a game we have that:

$P(x\geq 1)=1-P(x=0)$

Now lets consider the a game in which no pairs are obtained. The first card can be any, so we have 12 out 12 cards. The second card can't be the one matching the first card, so we have 10 out of 11. Similarly for the 3rd and 4th cards, we obtain:

$P(x=0)=\frac{12}{12} \frac{10}{11} \frac{8}{10} \frac{6}{9} =\frac{16}{33}$

This results in

$P(x\geq 1)=1-\frac{16}{33}=\frac{17}{33}$