Bill has a set of 6 black cards and a set of 6 red cards : Problem Solving (PS)

My Profile Logout

Test's Subscription Expires:

My Follow Feed Kudos My Error Log

Messages and Replies

Settings Mark All Read

See All

Applicant Notifications

Settings Mark All Read

See All

Global notifications

Settings Mark All Read

See All

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

Sort by Date

Sort by Kudos

Author

Message

TAGS:

Hide Tags

Senior Manager

Joined: 12 Mar 2013

Posts: 251

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

Show Tags

Updated on: 27 Feb 2018, 08:03

00:00

Difficulty:

95% (hard)

Question Stats:

21% (02:46) correct

79% (03:07) wrong

based on 101 sessions

HideShow timer Statistics

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

_________________

We Shall Overcome... One day...

Originally posted by nahid78 on 27 Feb 2018, 07:13.
Last edited by nahid78 on 27 Feb 2018, 08:03, edited 1 time in total.

CEO

Joined: 11 Sep 2015

Posts: 3122

Location: Canada

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

Show Tags

27 Feb 2018, 07:37

Top Contributor

nahid78 wrote:

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
_________________

Test confidently with gmatprepnow.com

Math Expert

Joined: 02 Sep 2009

Posts: 50627

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

Show Tags

27 Feb 2018, 09:08

nahid78 wrote:

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

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager

Joined: 12 Mar 2013

Posts: 251

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

Show Tags

27 Feb 2018, 10:06

Bunuel wrote:

nahid78 wrote:

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.

_________________

We Shall Overcome... One day...

Senior Manager

Joined: 12 Mar 2013

Posts: 251

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

Show Tags

02 Mar 2018, 21:04

GMATPrepNow wrote:

nahid78 wrote:

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

_________________

We Shall Overcome... One day...

Intern

Joined: 10 Dec 2016

Posts: 13

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

Show Tags

05 Mar 2018, 07:30

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..

Senior Manager

Joined: 12 Mar 2013

Posts: 251

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

Show Tags

05 Mar 2018, 08:01

sekharm2389 wrote:

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

Bunuel, Please help me(us) out
_________________

We Shall Overcome... One day...

Intern

Joined: 27 May 2015

Posts: 7

Schools: ISB '18

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

Show Tags

30 Mar 2018, 10:10

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