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

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Manager
Joined: 29 Oct 2009
Posts: 196
GMAT 1: 750 Q50 V42
A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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Updated on: 11 Feb 2012, 02:26
3
10
00:00

Difficulty:

85% (hard)

Question Stats:

53% (02:33) correct 47% (02:24) wrong based on 270 sessions

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

Happy solving!

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http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Originally posted by sriharimurthy on 15 Nov 2009, 04:38.
Last edited by Bunuel on 11 Feb 2012, 02:26, edited 1 time in total.
Edited the question and added the OA
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Joined: 02 Sep 2009
Posts: 52254

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15 Nov 2009, 05:20
20
5
sriharimurthy 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?

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

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

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

4) $$(1/3)^2*(1/2)$$

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

Happy solving!

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 ($$\frac{1}{3^4}$$).
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##### General Discussion
Manager
Joined: 29 Oct 2009
Posts: 196
GMAT 1: 750 Q50 V42

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15 Nov 2009, 05:55
2
1
Having Bunuel post is as good as posting the OA...

In fact, better.. because not only are the answers always right but the explanations always perfect!

+1 to you Bunuel! (I think soon you'll have more kudos than posts!)

Cheers!
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Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Manager
Joined: 13 Aug 2009
Posts: 176
Schools: Sloan '14 (S)

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15 Nov 2009, 08:11
2
Good problem.

The first day the man can wear anything:
(Shoes)*(Shirts)*(Pants)
(2/2)*(3/3)*(3/3)

The second day the man must wear the same pair of shoes and not wear the shirt and pants from the day before:
(1/2)*(2/3)*(2/3)

The final day, the man must continue to wear the same pair of shoes and wear the one shirt and one pants that were not worn:
(1/2)*(1/3)*(1/3)

So multiplay all together to get the probability of this happening each consecutive day:

(2/2)*(3/3)*(3/3)*(1/2)*(2/3)*(2/3)*(1/2)*(1/3)*(1/3)=
(1/3)^4

Manager
Joined: 24 Jul 2009
Posts: 67
Location: United States
GMAT 1: 590 Q48 V24

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15 Nov 2009, 10:37
h2polo,

Kudos to you as well, u have been ruling the quant section, soon i see u becoming the moderator.
But keep in mind, u r competing with the best....Bunuel.
Manager
Joined: 13 Aug 2009
Posts: 176
Schools: Sloan '14 (S)

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16 Nov 2009, 02:51
Oh no... there is no competing with Bunuel. He solves the unsolvable. Hahaha...
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Posts: 338
Location: PDX

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18 Nov 2009, 15:53
I missed out on the activities in this forum lately and it's again a joy to read Bunuel's answers. Kudos
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Joined: 01 Nov 2009
Posts: 14
Re: probability of wearing dress  [#permalink]

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05 Apr 2010, 04:02
My question is:

day 2: probability of choosing shoe should be 1 as we can not choose one of the two shoes rather the one that was already chosen on day 1
same for day 3
Manager
Joined: 10 Aug 2009
Posts: 120
Re: probability of wearing dress  [#permalink]

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05 Apr 2010, 07:44
san03: On day 2 (and 3), the probability for the shoes should be 1/2
This is because given a shoe was chosen on day 1, the probability of choosing that same shoe the next day is 1/2.
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Posts: 14
Re: probability of wearing dress  [#permalink]

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16 Apr 2010, 00:15
Aah!! i think I got confused between number of ways and probability.
probability of choosing same shoe= 1C1/2C1
Many thanks
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Re: A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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13 Nov 2013, 11:27
2
Another way to look at the problem :-
!. Since a single pair of shoes has to be chosen and worn for all the three days,
No. of ways of selecting one pair of shoes from the available two = 2C1 = 2.
2. Now the remaining task for us is to choose different shirts and pants for the three days : -
1st day = 3 X 3
2nd day = 2 X 2
3rd day = 1 X 1
Multiplying all the terms
2C1 X 2 X 2 X 3 X 3 X 1 X 1 (Numerator)

now what will come in the denominator?
day Pants Pair of shoes Shirts
1st 3 2 3
2nd 3 2 3
3rd 3 2 3

That is = 3^3 X 2^3 X 3^3 (Denominator)

When the Numerator is divided by the Denominator, you will get the required answer = 1/(3)^4.
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Re: A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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23 Jul 2014, 04:47
sriharimurthy 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. $$5*(\frac{1}{3})^2$$

Happy solving!

1st Day - $$\frac{3}{3}$$ shirts $$*$$ $$\frac{2}{2}$$ shoes $$*$$ $$\frac{3}{3}$$ pants

2nd Day - $$\frac{2}{3}$$ shirts, $$*$$ $$\frac{1}{2}$$ shoes $$*$$ $$\frac{2}{3}$$ pants

3rd Day - $$\frac{1}{3}$$ shirts, $$*$$ $$\frac{1}{2}$$ shoes $$*$$ $$\frac{1}{3}$$ pants

Answer is $$(\frac{1}{3})^4$$
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A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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05 Oct 2015, 07:59
1
the probability of choosing a unique set of attire on the fisrt day = 1.
from the second day, if I choose to wear the same pair of shoes(s1) on second and third day, then the probaility = [(1/2)*(2/3)*(2/3)]*[(1/2)*(1/3)*(1/3)]= 1/81.

Similarly,I can choose to wear shoes(S2) on all three days. Here as well the probability = 1/81.

Hence, the total probaility = 1/81 + 1/81= 2/81
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A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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23 Apr 2018, 09:20
sriharimurthy 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. $$5*(\frac{1}{3})^2$$

Happy solving!

3 different shirts, 2 different pair of shoes, and 3 different pants..
For first morning, Probability of selecting 1 shirt, 1 pair of shoes, and 1 pair of pants = 3/3 * 2/2 *3/3 =1
For 2nd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 2/3 * 1/2 * 2/3 = 2/9 (2 fresh shirts left and 2 fresh pair of pants left)
For 3rd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 1/3 * 1/2 * 1/3 = 1/18 (1 fresh shirt left and 1 fresh pair of pants left)

So, probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated = 1 * 2/9 * 1/18 = $$\frac {1}{3^4}$$

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