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Bill has a set of 6 black cards and a set of 6 red cards. Each card ha
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03 Apr 2021, 00:51
03 Apr 2021, 00:51
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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 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) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card ha
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04 Apr 2021, 07:50
04 Apr 2021, 07:50
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GeminiHeat 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 at least one pair of cards that have the same value?
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
Number of ways of selecting \(4\) cards from \(12\) \(= (12)(11)(10)(9) = 11880\)
Number of pairs we have, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
No pair \(= (12)(10)(8)(6) = 5760\)
Probability (At-least one pair) \(= 1 - (No Pair) = 1 - \frac{(12)(10)(8)(6)}{(12)(11)(10)(9)} = 1 - \frac{16}{33} = \frac{17}{33}\)
Hence, option C
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card ha
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05 Apr 2021, 05:58
05 Apr 2021, 05:58
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GeminiHeat 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 at least one pair of cards that have the same value?
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
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
= 16/33
So, P(at least one pair) = 1 - 16/33
= 17/33
Answer: C
Cheers,
Brent
