Problem Solving

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, there are 2 cards in the deck that have the same 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?


8/33

62/165

17/33

103/165

25/33

I got C as the ans.
Ways of selecting 4cards out of 12= 12*11*10*9...(X)

We will calculate probability of fetching cards with no 2 cards having same number.

This can be done in 12C1*10C1*8C1*6C1...(Y) (1st card from 12, 2nd from 10 as 1st card and other card having same no r out, 3rd card out of 8as 2 cards and their pair totalling 4 are out, 4th card from 6 as 3 cards and their pair totalling 6 are out)

Therefore, probability of getting no pair is Y/X = 16/33

Hence, for atleast 1 pair, probibility is 17/33.

Let me know, if my thinking is wrong.