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Bill has a set of 6 black cards and a set of 6 red cards

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Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post Updated on: 27 Feb 2018, 09:03
1
7
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

21% (02:54) correct 79% (03:08) wrong based on 104 sessions

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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 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
Spoiler: OA

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Originally posted by nahid78 on 27 Feb 2018, 08:13.
Last edited by nahid78 on 27 Feb 2018, 09:03, edited 1 time in total.
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 27 Feb 2018, 08:37
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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
= 16/33

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

Cheers,
Brent
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 27 Feb 2018, 10:08
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


This is a copy of the following MGMAT question: https://gmatclub.com/forum/bill-has-a-s ... 57417.html
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 27 Feb 2018, 11:06
Bunuel wrote:
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


This is a copy of the following MGMAT question: https://gmatclub.com/forum/bill-has-a-s ... 57417.html


Actually i have changed it a little to understand more deeply; the change is marked RED. :)
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 02 Mar 2018, 22:04
GMATPrepNow wrote:
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
= 16/33

So, 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.
Thank you :)
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 05 Mar 2018, 08:30
This question asks about exactly one pair and not atleast one pair. So, below is my solution

first place can have any of the 12 cards,
Out of 2nd , 3rd and 4th, only one will match with the first and rest two will be from different pairs.
12/12 * 1/11 *10/10 * 8/9 * 3c1 = 8/33

Please correct me if i m wrong..
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Re: Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 05 Mar 2018, 09:01
sekharm2389 wrote:
This question asks about exactly one pair and not atleast one pair. So, below is my solution

first place can have any of the 12 cards,
Out of 2nd , 3rd and 4th, only one will match with the first and rest two will be from different pairs.
12/12 * 1/11 *10/10 * 8/9 * 3c1 = 8/33

Please correct me if i m wrong..


Thank you,
but Why 3c1, answer is 16/33

Bunuel, Please help me(us) out
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Bill has a set of 6 black cards and a set of 6 red cards  [#permalink]

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New post 30 Mar 2018, 11:10
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ANSWER: B
Ways to select one pair out of 6 pairs: 6C1=6
Ways to select 2 cards out of remaining 10 cards such that no pair is drawn: 10C2-5. Please note 5 has to be subtracted for possible 5 pairs that were included in 10C2.
Required probability: (6*(10C2-5))/12C4 =(6*(45-5))/12C4=16/33

Alto:
One pair can be selected out of 6 pairs in 6C1 ways.
Probability of 1 pair thus selected and 2 different number cards: (12/12*1/11*10/10*8/9)=8/9*11
Probability of all six pairs: 6*8/9*11=16/33

Alto:
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

Now, P(at least one pair)=P(exactly one pair)+P(two pairs)
P(two pairs)= 6C2/12C4=1/33
Now, P(exactly one pair)=17/33-1/33=16/33
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Bill has a set of 6 black cards and a set of 6 red cards   [#permalink] 30 Mar 2018, 11:10
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