nahid78 wrote:

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 exactly one pair of cards that have the same value?

(A) 8/33

(B) 16/33

(C) 17/33

(D) 1/33

(E) 20/33

We can solve this question using

probability rules.

First, recognize that P(at least one pair) = 1 -

P(no pairs)P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)

= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)

= 1 x 10/11 x 8/10 x 6/9

=

16/33So, P(at least one pair) = 1 -

16/33= 17/33

= C

Cheers,

Brent

_________________

Brent Hanneson – GMATPrepNow.com

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