Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Aug 2016, 17:09

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A man chooses an outfit from 3 different shirts, 2 different

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

SVP
Joined: 21 Jul 2006
Posts: 1538
Followers: 10

Kudos [?]: 634 [4] , given: 1

A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

28 Nov 2007, 03:59
4
This post received
KUDOS
5
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

45% (03:16) correct 55% (02:23) wrong based on 166 sessions

### HideShow timer Statistics

A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A. $$(\frac{1}{3})^6*(\frac{1}{2})^3$$

B. $$(\frac{1}{3})^6*(\frac{1}{2})$$

C. $$(\frac{1}{3})^4$$

D. $$(\frac{1}{3})^2*(\frac{1}{2})$$

E. $$5*(\frac{1}{3})^2$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-man-chooses-an-outfit-from-3-different-shirts-2-different-91717.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2013, 07:59, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Manager
Joined: 20 Jun 2007
Posts: 157
Followers: 1

Kudos [?]: 16 [0], given: 0

[#permalink]

### Show Tags

28 Nov 2007, 04:27
I don't know if I'm right, but this is how I approached it:

Shirts Shoes Pants
Day 1 3/3 2/2 3/3
(It doesn't matter what he chooses on day 1)
Day 2 2/3 1/2 2/3
(On day 2 he can't choose the shirt or pants he had on Day 1, so there's a 2/3 chance of getting each of those right. And he has to choose the same shoes, a 1/2 chance)
Day 3 1/3 1/2 1/3
(On day 3 he can't choose the shirt or pants he had on Day 1 or 2, so there's a 1/3 chance of getting each of those right. And he has to choose the same shoes, a 1/2 chance)

Multiply all probabilities and you get 1*2/9*1/18 = 1/81 or 1/3^4
Answer:(C)
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 130 [0], given: 2

[#permalink]

### Show Tags

28 Nov 2007, 05:51
i sort of approached it the same way as above, but instead of having 2/3 for shirt and pants on day 2, i had 1/2

my logic was on day 2 he only has 2 shirts and 2 pants to pick from, and he picks one .... i didnt end up with any of those answer choices though
SVP
Joined: 21 Jul 2006
Posts: 1538
Followers: 10

Kudos [?]: 634 [0], given: 1

[#permalink]

### Show Tags

28 Nov 2007, 08:48
the OA is C, however, I don't even know how to approach this, so while i'm trying to figure this out, you guys can also think of the appropriate steps in approaching it. anyone managed to figure it out? Although Raffie got the correct answer, i'm really not too sure of the steps that you have taken. or is it appropriate?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 500

Kudos [?]: 3056 [0], given: 360

[#permalink]

### Show Tags

28 Nov 2007, 12:02
C.

we have

t=(2/2*1/2*1/2) - for shoes.
q=(3/3*2/3*1/3) - for shirts and the same fore pants.

P=t*q*q=(2/2*1/2*1/2)*(3/3*2/3*1/3)*(3/3*2/3*1/3)=1/4*4/3^4=(1/3)^4
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 373 [0], given: 0

[#permalink]

### Show Tags

28 Nov 2007, 12:49
walker wrote:
C.

we have

t=(2/2*1/2*1/2) - for shoes.
q=(3/3*2/3*1/3) - for shirts and the same fore pants.

P=t*q*q=(2/2*1/2*1/2)*(3/3*2/3*1/3)*(3/3*2/3*1/3)=1/4*4/3^4=(1/3)^4

Aha. Very nice!!!

I forgot about the 1/3 on the last day! I kept making it 2/3. Makes so much sense now!

Set it up exactly as you did cept for the 1/3, which made all the difference!
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 130 [0], given: 2

[#permalink]

### Show Tags

16 Dec 2007, 06:20
had to come back to this one and ask:

why is probability on day2 for pants and shirts 2/3 ?

if we are told that these items cannot repeat, then since one item was already selected on day 1, doesnt that only leave 2 items for him to choose from on day2 ?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 500

Kudos [?]: 3056 [0], given: 360

[#permalink]

### Show Tags

16 Dec 2007, 06:37
Expert's post
1
This post was
BOOKMARKED
pmenon wrote:
had to come back to this one and ask:

why is probability on day2 for pants and shirts 2/3 ?

if we are told that these items cannot repeat, then since one item was already selected on day 1, doesnt that only leave 2 items for him to choose from on day2 ?

You are right in the case without replacement. But "he wears the same pair of shoes each day" means replacement for shoes. So we can assume that there also is replacement for pants and shirts and he can wear of 3 items of pants and shirts on 2nd an 3rd day.
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 130 [0], given: 2

[#permalink]

### Show Tags

16 Dec 2007, 09:48
Thanks, walker ! I guess I had overlooked that .

Yes absolutely, if there is replacement, there are always 3 choices to select from for pants and shirt. Good call !
Manager
Joined: 01 Sep 2007
Posts: 100
Location: Astana
Followers: 1

Kudos [?]: 22 [0], given: 0

[#permalink]

### Show Tags

22 Dec 2007, 23:53
P (3 different shirts) = 1*2/3*1/3
P (same shoes) = 1/2*1/2
P (3 different pants) = 1*2/3*1/3

(1/9)^2 = (1/3)^4
Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 622 [1] , given: 33

Re: PS: Probability [#permalink]

### Show Tags

26 Aug 2008, 23:30
1
This post received
KUDOS
Total number fo ways = (3*2*3) = 18

Day one: 3*2*3/18 = 1
Day two: 2*1*2/18 = 4/18
Day three: 1/18 = 1/18

It is important to nore that we multply the probabilties of all three days, in order to get the total probability because selection of same things on different days from the same set makes the events as dependent events.

Please see the attached picture for more clarity.

(2*3^2)*(2^2)*(1)/(3^6*2^3) = [(2^3)*(3^2)]/[(2^3)*(3^6)] = (1/3)^4
Attachments

File comment: Probability

Probability.PNG [ 40.59 KiB | Viewed 4870 times ]

Manager
Joined: 22 Jul 2008
Posts: 154
Followers: 1

Kudos [?]: 12 [0], given: 0

Re: PS: Probability [#permalink]

### Show Tags

27 Aug 2008, 14:30
Day#1 Day#2 Day#3
Shirts: 3 x 2 x 1

Pants: 3 x 2 x 1

Shoes: 2 x 1 x 1

Total probable outcomes = 6 * 6 * 2 = 72

Total possible outcomes = 2^3 * 3^3 * 3^3

Therefore, Prob. = 72/2^3*3^3*3^3 =1/3^4
Current Student
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Followers: 1

Kudos [?]: 46 [0], given: 155

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

17 Jul 2013, 07:52
I was wondering how to solve this problem with the combinatoric approach. But I can´t figure it out.
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Math Expert
Joined: 02 Sep 2009
Posts: 34449
Followers: 6271

Kudos [?]: 79582 [1] , given: 10022

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

17 Jul 2013, 08:02
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
Maxirosario2012 wrote:
I was wondering how to solve this problem with the combinatoric approach. But I can´t figure it out.

Better to use probability approach for this question:
A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A. $$(1/3)^6*(1/2)^3$$

B. $$(1/3)^6*(1/2)$$

C. $$(1/3)^4$$

D. $$(1/3)^2*(1/2)$$

E. $$5*(1/3)^2$$

For the first day he can choose any outfit, $$p=1$$;

For the second day he must choose the same shoes as on the first day and different shirts and pants form the first day's, $$p=\frac{1}{2}*\frac{2}{3}*\frac{2}{3}=\frac{2}{9}$$;

For the third day he must choose the same shoes as on the first day and different shirts and pants from the first and second day's, $$p=\frac{1}{2}*\frac{1}{3}*\frac{1}{3}=\frac{1}{18}$$;

$$P=1*\frac{2}{9}*\frac{1}{18}=\frac{1}{81}=\frac{1}{3^4}$$

Answer: C ($$\frac{1}{3^4}$$).

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-man-chooses-an-outfit-from-3-different-shirts-2-different-91717.html
_________________
Current Student
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Followers: 1

Kudos [?]: 46 [0], given: 155

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

17 Jul 2013, 08:13
Thank you Bunuel. Finally I solved it using probabilities. But I think there must be a way to solve this problem using combinatorics. But it is too difficult to imagine, at least for me.
Obviusly, in the test I will try with probability to solve this kind of problems. But I like to solve all this problems in different ways, just for practice.
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Intern
Joined: 08 Aug 2013
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

08 Aug 2013, 04:43
I get the Numerator part - Probable outcomes. Cannot figure out about the denominator - Possible outcomes. (referring to August 2008 post by Kassalmd) Can someone help?
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 173

Kudos [?]: 1795 [0], given: 219

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

08 Aug 2013, 05:41
itikakulkarni wrote:
I get the Numerator part - Probable outcomes. Cannot figure out about the denominator - Possible outcomes. (referring to August 2008 post by Kassalmd) Can someone help?

The denominator must represent the total number of combinations, according to the formula P=favorableOutcome/TotalOutcomes.

He can choose from 3 different shirts, 2 different pairs of shoes, and 3 different pants each day.

So the total number of outfits with the shirts is 3*3*3 (each day 3 possibilities), the total number of shoes is 2*2*2, and the one for pants is 3*3*3 (with the same logic). So the denominator= Total Outcomes = $$2^3 * 3^3 * 3^3$$.

Hope it's clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Senior Manager
Joined: 10 Jul 2013
Posts: 335
Followers: 3

Kudos [?]: 275 [0], given: 102

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]

### Show Tags

10 Aug 2013, 15:25
tarek99 wrote:
A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A. $$(\frac{1}{3})^6*(\frac{1}{2})^3$$

B. $$(\frac{1}{3})^6*(\frac{1}{2})$$

C. $$(\frac{1}{3})^4$$

D. $$(\frac{1}{3})^2*(\frac{1}{2})$$

E. $$5*(\frac{1}{3})^2$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-man-chooses-an-outfit-from-3-different-shirts-2-different-91717.html

Gist: Every day shirt and pant decreases by 1 to choose and shoe remains the same one
1st day = 3/3 * 2/2 * 3/3 = 1
2nd day = 2/3 * 1/2 * 2/3 = 2/9
3rd day = 1/3 * 1/2 * 1/3 = 1/18
so required probability = 1 * 2/9 * 1/18 = 1/81 = 1/3^4
_________________

Asif vai.....

Re: A man chooses an outfit from 3 different shirts, 2 different   [#permalink] 10 Aug 2013, 15:25
Similar topics Replies Last post
Similar
Topics:
8 If the number of different positive factors of (2^y)(3^3) is the same 5 13 Apr 2015, 09:43
5 If y*(3x - 5)/2 = y and y is different from 0, then x = 6 22 Dec 2010, 17:28
1 A certain businessman wears only three different colors of shirt -- wh 3 16 Dec 2010, 01:32
6 In how many different ways can 3 identical green shirts and 10 25 Nov 2010, 18:54
32 A man chooses an outfit from 3 different shirts, 2 different 14 15 Nov 2009, 05:38
Display posts from previous: Sort by

# A man chooses an outfit from 3 different shirts, 2 different

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.