a beautiful pair !

[rohilla_sandeep]

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?

8/33

62/165

17/33

103/165

25/33

Answer is 'C'

explanation : Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 - 16/33 = 17/33.


Doubt : I will take an example to show that explanation is wrong. Consider three cards : A1, A2 & B1 where A1 and A2 are similar but of different suits. Therefore, according to explanation the chances of getting atleast one pais is = 1 - (1)*(1/2) = 1/2 but actually is should be 1/3 (A1A2, A1b1, a2b2).

[jnelson0612]

rohilla, please read this thread about this problem and post if you have further questions: bill-has-a-small-deck-of-12-playing-cards-t1898.html