GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 02 Jun 2020, 07:10

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

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

Emily keeps 12 different pairs of shoes (24 individual shoes in total)

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 64168
Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

16 Sep 2018, 22:20
10
00:00

Difficulty:

25% (medium)

Question Stats:

70% (01:13) correct 30% (01:25) wrong based on 208 sessions

HideShow timer Statistics

Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

_________________
Intern
Joined: 26 Aug 2018
Posts: 6
Location: Mexico
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

17 Sep 2018, 12:10
1
Probability of the first one don't not matter, so we should consider what is the probability of the second one be the pair of the first one.

1/23

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10640
Location: United States (CA)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

18 Sep 2018, 17:50
1
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

The first shoe can be any shoe, so its probability is 24/24 = 1. However, since the second shoe must match the first shoe, its probability of being chosen is 1/23. Therefore, the probability of the two shoes forming a matching pair is 1 x 1/23 = 1/23.

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
202 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 937
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

09 Jan 2019, 08:05
1
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

$$\left. \matrix{ {\rm{Total}}\,\,:\,\,\,C\left( {24,2} \right) = {{24 \cdot 23} \over 2} = 12 \cdot 23\,\,{\rm{equiprobable}}\,\,{\rm{pairs}}\,\,\,\, \hfill \cr {\rm{Favorable:}}\,\,\,12\,\,{\rm{real}}\,\,{\rm{pairs}}\,\,\left( {{\rm{matches}}} \right) \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {{12} \over {12 \cdot 23}} = {1 \over {23}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4043
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

15 Apr 2019, 03:05
1
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

Method 1:

Probability = Favourable outcomes / Total outcomes = Total Pairs / Total ways of picking 2 shoes out of 24

i.e. Required probability $$= \frac{12C1}{24C2} = \frac{12}{(12*23)} = \frac{1}{23}$$

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability $$= \frac{(24*1)}{(24*23)}$$

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting
Check website for most affordable Quant on-Demand course 2000+ Qns (with Video explanations)
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10640
Location: United States (CA)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

16 Apr 2019, 16:55
1
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

The probability that the dog will drag out any matching pair is:

2/24 x 1/23 = 1/12 x 1/23

Since there are 12 different pairs, the total probability is 12 x (1/12 x 1/23) = 1/23.

Alternate Solution:

The dog can drag any shoe out, leaving 23 shoes under the bed. The probability that the next shoe he drags out is its match is 1/23.

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
202 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4043
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

24 Sep 2019, 18:52
1
ghnlrug wrote:
GMATinsight wrote:

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability $$= \frac{(24*1)}{(24*23)}$$

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

why do you not have to divide both (24*1) and (24*23) by 2? Or did you just leave that out?

ghnlrug

In probability calculation, you must calculate both numerator and denominator either by selection (Method 1) or by Arrangement (Method 2) so in either case, you get the correct answer therefore dividing by 2 wasn't needed.
_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting
Check website for most affordable Quant on-Demand course 2000+ Qns (with Video explanations)
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
Current Student
Joined: 24 Aug 2016
Posts: 788
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

17 Sep 2018, 12:04
2
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

No of ways selecting the 1st shoe = 24
No of ways selecting the 1st & 2nd shoe= 24 *23
No of ways selecting the 2nd shoe which is the exact pair of the 1st = 1
No of ways selecting selecting the 1st shoe & the 2nd shoe which is the exact pair of the 1st= 24*1
Hence P = 24*1 /24*23 = 1/23 ....... Ans C
_________________
Please let me know if I am going in wrong direction.
Thanks in appreciation.
Manager
Joined: 23 Aug 2016
Posts: 101
Location: India
Concentration: Finance, Strategy
GMAT 1: 660 Q49 V31
GPA: 2.84
WE: Other (Energy and Utilities)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

17 Sep 2018, 20:11
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

Hi,

My approach-

Probability of selecting a particular shoe from the 12 different pair=2/24
Now probability of selecting the matching shoe to the shoe already dragged= 1/23(remember one shoe is already out)

Therefore combined probability of one matching pair of shoe= 1/24* 1/23

Now, there are 12 different pair of shoe so your probability will shoot up 12 times-

2/24 *1/23 *12=1/23.

Bunuel is it correct?
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4882
GMAT 1: 770 Q49 V46
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

09 Jan 2019, 06:15
Top Contributor
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

P(dog selects matching pair) = P(dog chooses ANY sock 1st AND 2nd sock matches the 1st sock)
= P(dog chooses ANY sock 1st) x P(2nd sock matches the 1st sock)
= 24/24 x 1/23
= 1/23

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Senior Manager
Joined: 03 Sep 2018
Posts: 259
Location: Netherlands
Schools: HEC MiM "22 (S)
GPA: 4
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

24 Sep 2019, 01:38
Ways to select first = 24
Ways to select second = 1
But since the order is irrelevant, there are really just 12 ways
Total ways to choose 2 from 24 (order irrelevant) = 24C2

hence 12/24C2
_________________
Good luck to you. Retired from this forum.
Senior Manager
Joined: 03 Sep 2018
Posts: 259
Location: Netherlands
Schools: HEC MiM "22 (S)
GPA: 4
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

24 Sep 2019, 01:40
GMATinsight wrote:

Method 2:

Probability = Favourable outcomes / Total outcomes = Total ways of picking two shoes making pair / Total ways of picking 2 shoes out of 24

i.e. Required probability $$= \frac{(24*1)}{(24*23)}$$

Favourable outcomes: first shoe may be picked in 24 ways but second shoe may be picked only in one way to make pair of shoe
Total outcomes: first shoe may be picked in 24 ways and second shoe may be picked in 23 ways

why do you not have to divide both (24*1) and (24*23) by 2? Or did you just leave that out?
_________________
Good luck to you. Retired from this forum.
CEO
Joined: 03 Jun 2019
Posts: 2921
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)  [#permalink]

Show Tags

24 Sep 2019, 19:01
Bunuel wrote:
Emily keeps 12 different pairs of shoes (24 individual shoes in total) under her bed. If her dog drags out two shoes at random, what is the probability that he drags out a matching pair of shoes?

A. 1/144
B. 1/66
C. 1/23
D. 1/12
E. 1/11

Probability = 11*1/24*23 = 1/23

IMO C

Posted from my mobile device
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
Re: Emily keeps 12 different pairs of shoes (24 individual shoes in total)   [#permalink] 24 Sep 2019, 19:01