Last visit was: 23 Apr 2026, 12:46 It is currently 23 Apr 2026, 12:46
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
nahid78
User avatar
Joined: 12 Mar 2013
Last visit: 30 Apr 2023
Posts: 283
Own Kudos:
737
 [27]
Given Kudos: 1,062
Products:
My Reviews
Posts: 283
Kudos: 737
 [27]
Bill has a set of 6 black cards and a set of 6 red cards Updated on: 27 Feb 2018, 09:03
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.
3
Kudos
Add Kudos
24
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

27% (02:48) correct 73% (03:03) wrong based on 208 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
B
Most Helpful Reply
avatar
davezhcoet
avatar
Joined: 27 May 2015
Last visit: 03 Jul 2019
Posts: 5
Own Kudos:
15
 [11]
Given Kudos: 143
Schools: ISB '18
Schools: ISB '18
Posts: 5
Kudos: 15
 [11]
Bill has a set of 6 black cards and a set of 6 red cards 30 Mar 2018, 11:10
7
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
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
General Discussion
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,451
 [4]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,451
 [4]
Bill has a set of 6 black cards and a set of 6 red cards 27 Feb 2018, 08:37
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
nahid78
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
Show more

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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,847
Given Kudos: 105,853
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,847
Bill has a set of 6 black cards and a set of 6 red cards 27 Feb 2018, 10:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nahid78
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
Show more

This is a copy of the following MGMAT question: https://gmatclub.com/forum/bill-has-a-s ... 57417.html
avatar
kelbyandrews
avatar
Joined: 18 May 2021
Last visit: 02 Dec 2021
Posts: 14
Given Kudos: 4
Posts: 14
Kudos: 0
Bill has a set of 6 black cards and a set of 6 red cards 26 Jun 2021, 08:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nahid78
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
Show more


Everyone is forgetting to account for the possibility of two pairs. Everyone says "P(exactly one pair) = 1 - P(No pairs). Does anyone know how to do it with P(Exactly one pair) = 1 - (P(no pairs)*P(2 pairs)?

This is my math.

P(No Pairs) = (12/12) * (10/11) * (8/10) * (6/9) = 16/33
P(Two Pairs) = (12/12) * (1/11) * (10/10) * (1/9) = 1/99

Then I multiplied them both and got roughly 1/198.

I feel like everyone forgets the possibility of there being two pairs. Taking that into account, is this correct?

**EDIT**
I retried this problem. Here is my new math.

"Probability of exactly 1 pair" is similar to getting A - A - B - C in which A,B & C are all digits 1-6.
So the Probability of getting A or any number for that matter is (1)
Probability of matching that same number A is (1/11)
Probability of getting any other number B is (1)
Probability of getting any other number that doesn't match B is C = (8/9)
Hence, (1)*(1/11)*(1)*(8/9)
Then you need to account for all the possible permutations of A - A - B - C which is (4!/2!)
Therefore, Probability of getting exactly 1 pair =
(1)*(1/11)*(1)*(8/9)*(4)*(3) = 8/11

Please let me know if this is correct!!
User avatar
ThebladeRunner
User avatar
Joined: 24 Dec 2022
Last visit: 22 Sep 2023
Posts: 3
Own Kudos:
2
 [1]
Given Kudos: 4
Posts: 3
Kudos: 2
 [1]
Bill has a set of 6 black cards and a set of 6 red cards 07 Sep 2023, 04:32
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Alternate Solution:

Ways to select one pair out of 6 possibilities (1-1,2-2,3-3,4-4,5-5,6-6): 6C1
Now the remaining 2 cards cannot be a pair, so the 1st of them will have 5C1 possibility (since one number is already taken up in the pair cards).
2nd non-pair card will have 4C1 possibility (as 1 is gone with pair card, and it cannot be same as the 1st non-pair card)

But, you also need to consider for colors, so non-pair numbers can have reverse so we multiply by 2.
Hence total ways: 6C1 x 5C1 x 4C1 x 2
And total possibilities: 12C4

Answer: (6C1 x 5C1 x 4C1 x 2) / 12C4 = 16/33
User avatar
pudu
User avatar
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 229
Own Kudos:
123
Given Kudos: 16
Location: India
Posts: 229
Kudos: 123
Bill has a set of 6 black cards and a set of 6 red cards 06 Oct 2023, 17:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
please some one provide clearly how to fin out exactly one pair. don't confuse us
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
1,117
Posts: 38,964
Kudos: 1,117
Bill has a set of 6 black cards and a set of 6 red cards 01 Apr 2026, 11:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109785 posts
Krunaal
Tuck School Moderator
853 posts