Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, 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?
P(at least 1 pair of the SAME value) = 1 - P(selecting 4 DIFFERENT values).hemanthkumarmn wrote:Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, 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. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33
P(selecting 4 different values):
The first card selected can be ANY VALUE.
P(2nd card is a different value) = 10/11. (Of the 11 cards remaining, any but the mate of the 1st card, leaving 11-1 = 10 good options.)
P(3rd card is a different value) = 8/10. (Of 10 cards remaining, any but the mates of the first 2 cards, leaving 10-2 = 8 good options.)
P(4th card is a different value) = 6/9. (Of the 9 cards remaining, any but the mates of the first 3 cards, leaving 9-3 = 6 good options.)
Since we want all of these events to happen, we multiply the probabilities:
10/11 * 8/10 * 6/9 = 16/33.
Thus:
P(at least 1 pair of the same value) = 1 - 16/33 = 17/33.
The correct answer is C.
