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A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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28 Nov 2007, 03:59
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44% (03:15) correct
56% (02:16) wrong based on 181 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. \(5*(\frac{1}{3})^2\) OPEN DISCUSSION OF THIS QUESTION IS HERE: amanchoosesanoutfitfrom3differentshirts2different91717.html
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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.



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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)



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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



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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?



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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



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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!



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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 ?



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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.



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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 !



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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



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Re: PS: Probability [#permalink]
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26 Aug 2008, 23:30
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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
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Re: PS: Probability [#permalink]
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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



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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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17 Jul 2013, 07:52
I was wondering how to solve this problem with the combinatoric approach. But I can´t figure it out.
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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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17 Jul 2013, 08:02
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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: amanchoosesanoutfitfrom3differentshirts2different91717.html
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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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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.
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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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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?



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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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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
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Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
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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: amanchoosesanoutfitfrom3differentshirts2different91717.htmlGist: 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
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