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A man chooses an outfit from 3 different shirts, 2 different [#permalink]
15 Nov 2009, 04:38

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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?

Re: Snappy dresser.. :P [#permalink]
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, \(p=\frac{1}{2}*\frac{1}{3}*\frac{1}{3}=\frac{1}{18}\);

A man chooses an outfit from 3 different shirts, 2 different [#permalink]
05 Oct 2015, 07:59

1

This post received KUDOS

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.

Re: Snappy dresser.. :P [#permalink]
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.

Re: probability of wearing dress [#permalink]
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..

Re: probability of wearing dress [#permalink]
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.

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

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

Re: A man chooses an outfit from 3 different shirts, 2 different [#permalink]
04 Oct 2015, 05:22

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