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?

A.8/33
B.62/165
C.17/33
D.103/165
E.25/33

I came up with C

Total 12 cards (1,2,3,4,5,6 and 1,2,3,4,5,6)
Ways to pick 4 cards (12*11*8*9)
Atleast one pair = 1 - (No pair)

How to pick a non pair ? For the first card we have 12 choices (assume we picked 6). For the second card we have 10 choices (cant pick 6. Assume the second card picked was 3). Third card we have 8 choices (Cant pick 6 and 3. Assume third card picked was 5). Fourth card we have 6 choices (Cant pick 6 ,3 or 5 )


So this boils down to = (12*10*8*6)/(12*11*8*9) = 16/33

Answer = 1-(16/33) = 17/33

HTH

Can you please tell me why the ways to pick the cards is 12.11.8.9 and not 12C4

Because it doesn't say a group of 4 cards. Generally, whenever something is picked in a group, the question uses very explicit words (ex: group, team,committee,at once.. etc etc)

raulverde wrote: Ways to pick 4 cards (12*11*8*9)

This should read 12*11*10*9, right?

yes...12*11*10*9.

As you can tell i am not a big fan of proofreading