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Please help - probability question

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Please help - probability question [#permalink] New post 26 Nov 2006, 19:33
Can someone please explain to me the answer? This is from Princeton Review practice test:

Q: 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)^4
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 [#permalink] New post 26 Nov 2006, 19:45
first step:

work out the total #:

every day he can select any of them which are independent:
3 * 2 * 3.

and each day is also independent from each other, so for 3 days, you have:
(3*2*3)^3.
This is the total # that'll be the denominator.

Second step:
find the desired #:

Since shoes for 3 days don't change, so for shoes u have 2 choices for all 3 days:

2.

Shirts change every day and no repeat, it is a permutation of all the 3 shirts. so it is 3!.

pants are the same as shirts are. so also 3!.

shoes, shirts and pants are independent from one another, so multiply them:
2*3!*3!.

so the prob:

2*3!*3!/ (3*2*3)^3. u get the answer.
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Thanks again [#permalink] New post 26 Nov 2006, 20:07
I really appreciate the help :-D
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 [#permalink] New post 26 Nov 2006, 20:09
tennis_ball wrote:
first step:

work out the total #:

every day he can select any of them which are independent:
3 * 2 * 3.

and each day is also independent from each other, so for 3 days, you have:
(3*2*3)^3.
This is the total # that'll be the denominator.

Second step:
find the desired #:

Since shoes for 3 days don't change, so for shoes u have 2 choices for all 3 days:

2.

Shirts change every day and no repeat, it is a permutation of all the 3 shirts. so it is 3!.

pants are the same as shirts are. so also 3!.

shoes, shirts and pants are independent from one another, so multiply them:
2*3!*3!.

so the prob:

2*3!*3!/ (3*2*3)^3. u get the answer.


Shirts change every day and no repeat, it is a permutation of all the 3 shirts. so it is 3!.

Why are arranging shirts...You should be rather picking ie a combination should come right?

Where am I thinking wrong tennis?
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 [#permalink] New post 26 Nov 2006, 21:27
you can think either way.

if you think pick:

1st day: 3 choices to pick.
2nd day: 2 choices to pick.
3rd day: 1 choices.

3*2*1.

it is the same as thinking it as permutation of 3: Stick 3 different shirts onto the 3 days. :-D
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 [#permalink] New post 27 Nov 2006, 04:32
tennis_ball wrote:
you can think either way.

if you think pick:

1st day: 3 choices to pick.
2nd day: 2 choices to pick.
3rd day: 1 choices.

3*2*1.

it is the same as thinking it as permutation of 3: Stick 3 different shirts onto the 3 days. :-D


oops ..got it

Thanks dude :-D
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 [#permalink] New post 27 Nov 2006, 08:53
Day 1
shooes *shirt*trouser = (1)*(1)*(1) = 1 He can choose any shoe, shirt or trouser will do

Day 2 = (1/2)*(2/3)*(2/3) = 2/ (3^2)he has to choose same shoe and any of the other two trousers , shirts

Day 3 = (1/2)*(1/3)*(1/3) = 1/ (2*3^2)choose same shoe and the only other shirt and trouse left

Total 1*(2/3^2)*(1/2*3^2)= 1/3^4
  [#permalink] 27 Nov 2006, 08:53
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