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 at least one pair of cards that have the same value?
(A) \(\dfrac{8}{33}\)
(B) \(\dfrac{62}{65}\)
(C) \(\dfrac{17}{33}\)
(D) \(\dfrac{103}{165}\)
(E) \(\dfrac{25}{33}\)
Answer: C
Source: Manhattan GMAT
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....
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We can solve this question using probability rules.Gmat_mission wrote: ↑Sun Sep 12, 2021 9:34 amBill 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 at least one pair of cards that have the same value?
(A) \(\dfrac{8}{33}\)
(B) \(\dfrac{62}{65}\)
(C) \(\dfrac{17}{33}\)
(D) \(\dfrac{103}{165}\)
(E) \(\dfrac{25}{33}\)
Answer: C
Source: Manhattan GMAT
First, recognize that P(at least one pair) = 1  P(no pairs)
P(no pairs) = P(select ANY 1st card AND select any nonmatching card 2nd AND select any nonmatching card 3rd AND select any nonmatching card 4th)
= P(select any 1st card) x P(select any nonmatching card 2nd) x P(select any nonmatching card 3rd) x P(select any nonmatching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33
So, P(at least one pair) = 1  16/33
= 17/33
Answer: C
Cheers,
Brent