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
Thank you for your response. But I have changed the question a little and now i have no idea how to solve it. Any help will be appreciated.
We Shall Overcome... One day...