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A man chooses an outfit from 3 different shirts, 2 different
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Updated on: 11 Feb 2012, 03:26
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
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15 Nov 2009, 06:20
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?
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
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15 Nov 2009, 06:55
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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Re: Snappy dresser.. :P
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15 Nov 2009, 09:11
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
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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
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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
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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
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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
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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
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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
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13 Nov 2013, 12:27
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!
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Re: A man chooses an outfit from 3 different shirts, 2 different
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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
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05 Oct 2015, 08:59
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
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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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