pants shoes and shirt

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pants shoes and shirt

by Oneva » Sun Jan 09, 2011 11:50 am
A man has 3 shirts, 2 pairs of shoes and 3 pants. If he chooses one shirt one pair of shoes and one pant each day for 3 days. What is the probability that the man will choose the same pair of shoes for all 3 days? It does not matter what he chooses for shirt or pants.

I couldn't remember all the details of this question because it was an online practie test. I was wondering if someone would be able to solve this problem without the answer choices. I couldnt figure it out.

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by anshumishra » Sun Jan 09, 2011 12:51 pm
Oneva wrote:A man has 3 shirts, 2 pairs of shoes and 3 pants. If he chooses one shirt one pair of shoes and one pant each day for 3 days. What is the probability that the man will choose the same pair of shoes for all 3 days? It does not matter what he chooses for shirt or pants.

I couldn't remember all the details of this question because it was an online practie test. I was wondering if someone would be able to solve this problem without the answer choices. I couldnt figure it out.
If It does not matter what he chooses for shirt or pants. :

First Day -> wear anything, choose any shoe pair = 1
Second day -> wear anything, choose the same pair = 1/2
third day -> wear anything, choose the same pair = 1/2

So, probability = 1*(1/2)*(1/2) = 1/4

If clothing (shirt or pants) shouldn't be repeated :

First Day -> wear anything, choose any shoe pair = 1
Second day -> wear different shirt, different pant, choose the same pair of shoe = 2/3 * 2/3 *1/2 = 2/9
third day -> wear the last shirt left, wear the last pant, choose the same pair of shoe = 1/3* 1/3 *1/2 = 1/18

So, probability = 2/9*1/18 = 1/81
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Anshu

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by frank1 » Mon Jan 10, 2011 4:15 am
Oneva wrote:A man has 3 shirts, 2 pairs of shoes and 3 pants. If he chooses one shirt one pair of shoes and one pant each day for 3 days. What is the probability that the man will choose the same pair of shoes for all 3 days? It does not matter what he chooses for shirt or pants.

I couldn't remember all the details of this question because it was an online practie test. I was wondering if someone would be able to solve this problem without the answer choices. I couldnt figure it out.
I think this is manhattan's question.
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by GMATGuruNY » Mon Jan 10, 2011 4:31 am
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
This question is from the Princeton Review. Quoted above is the entire question, along with the answer choices.

The first day is immaterial; the man can wear whatever he wants. We're concerned only with what the man chooses to wear the second and third days. Here are the outcomes we need:

Second day: different shirt, different pair of pants, same pair of shoes
Third day: different shirt from those worn on days 1 and 2, different pair of pants from those worn on days 1 and 2, same pair of shoes


Let's determine each probability separately:

Probability of choosing a different shirt on the second day: 2/3 (because he has 3 shirts total, and he can't wear the same shirt worn the first day, leaving him 2 good choices)

Probability of choosing a different pair of pants on the second day: 2/3 (because he has 3 pairs total, and he can't wear the same pair worn the first day, leaving him 2 good choices)

Probability of choosing the same pair of shoes: 1/2 (because he has 2 pairs total, and he has to wear the same pair worn the first day, leaving him 1 good choice)

Probability of choosing a different shirt on the third day: 1/3 (because he has 3 shirts total, and he can't wear the same shirts worn the first and second days, leaving him only 1 good choice)

Probability of choosing a different pair of pants on the third day: 1/3 (because he has 3 pairs total, and he can't wear the same pairs worn the first and second days, leaving him only 1 good choice)

Probability of choosing the same pair of shoes: 1/2 (because he has 2 pairs total, and he has to wear the same pair worn the first day and second days, leaving him 1 good choice)

We need all of these events to happen together in order to get a good outcome. To determine the probability that multiple events will happen together, remember this rule:

Probability (A and B) = Probability (A) * Probability (B)

We multiply the probabilities because the more things we want to happen together, the smaller the probability, and when you multiply fractions, the result keeps getting smaller.

So let's take the probabilities that we determined above and multiply:

2/3 * 2/3 * 1/2 * 1/3 * 1/3 * 1/2 = (1/3)^4

The correct answer is C.


Please note that the explanation above might seem long, but the process is quick and easy if you know what to do!
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by anshumishra » Mon Jan 10, 2011 6:19 am
Just as a note :
There are two different problems discussed here. I would advise you to go to both of them.
My first solution is for the problem, which the poster posted here (answer = 1/4).
The second solution is for the problem GMATGuruNY later posted and solved (answer = 1/81).

The two problems are different, as in the second problem "clothing (shirt or pants) shouldn't be repeated".
Thanks
Anshu

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