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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] New post 15 Nov 2009, 04:38
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48% (02:30) correct 52% (01:49) wrong based on 128 sessions
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!
[Reveal] Spoiler: OA

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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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Re: Snappy dresser.. :P [#permalink] New post 15 Nov 2009, 05:20
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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] New post 15 Nov 2009, 05:55
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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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Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

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

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Re: Snappy dresser.. :P [#permalink] New post 15 Nov 2009, 08:11
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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] New post 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.
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Re: Snappy dresser.. :P [#permalink] New post 16 Nov 2009, 02:51
Oh no... there is no competing with Bunuel. He solves the unsolvable. Hahaha...
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Re: Snappy dresser.. :P [#permalink] New post 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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Re: probability of wearing dress [#permalink] New post 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
please explain..
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Re: probability of wearing dress [#permalink] New post 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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Re: probability of wearing dress [#permalink] New post 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] New post 26 Aug 2013, 01:53
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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink] New post 13 Nov 2013, 11:27
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] New post 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
Re: A man chooses an outfit from 3 different shirts, 2 different   [#permalink] 23 Jul 2014, 04:47
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