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

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

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New post Updated on: 11 Feb 2012, 03:26
3
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A
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C
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E

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  85% (hard)

Question Stats:

53% (02:32) correct 47% (02:22) wrong based on 267 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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Originally posted by sriharimurthy on 15 Nov 2009, 05:38.
Last edited by Bunuel on 11 Feb 2012, 03:26, edited 1 time in total.
Edited the question and added the OA
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Re: Snappy dresser.. :P  [#permalink]

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New post 15 Nov 2009, 06:20
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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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Re: Snappy dresser.. :P  [#permalink]

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New post 15 Nov 2009, 06:55
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Having Bunuel post is as good as posting the OA... :wink:

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

Word Problems Made Easy!
1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

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Re: Snappy dresser.. :P  [#permalink]

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New post 15 Nov 2009, 09:11
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

ANSWER: C.
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Re: Snappy dresser.. :P  [#permalink]

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New post 15 Nov 2009, 11:37
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.
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Re: Snappy dresser.. :P  [#permalink]

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New post 16 Nov 2009, 03:51
Oh no... there is no competing with Bunuel. He solves the unsolvable. Hahaha...
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Re: Snappy dresser.. :P  [#permalink]

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New post 18 Nov 2009, 16:53
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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Re: probability of wearing dress  [#permalink]

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New post 05 Apr 2010, 05:02
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
please explain..
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Re: probability of wearing dress  [#permalink]

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New post 05 Apr 2010, 08:44
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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Re: probability of wearing dress  [#permalink]

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New post 16 Apr 2010, 01:15
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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New post 13 Nov 2013, 12:27
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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.
Kudos for the answer! :-D
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Re: A man chooses an outfit from 3 different shirts, 2 different  [#permalink]

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New post 23 Jul 2014, 05:47
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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New post 05 Oct 2015, 08:59
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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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New post 23 Apr 2018, 10:20
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 = \(\frac {1}{3^4}\)

Answer C
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