## 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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### 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....

by Gmat_mission » Sun Sep 12, 2021 9:34 am

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## Global Stats

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) $$\dfrac{25}{33}$$

Source: Manhattan GMAT

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### Re: 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

by [email protected] » Sun Sep 12, 2021 12:17 pm
Gmat_mission wrote:
Sun Sep 12, 2021 9:34 am
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) $$\dfrac{25}{33}$$

Source: Manhattan GMAT
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
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33