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a probability problem, who can help? Thanks!

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by billzhao » Wed Feb 18, 2009 5:01 pm
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

Answer is


[/spoiler]
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by earth@work » Wed Feb 18, 2009 6:37 pm
is the OA 1/18; also some part of the question seems to be missing in the end...i guess it says none of the other clothes are repeated

1st day: P1=1/3*1/3 =1/9
2nd day: P2=1/2*1/2 =1/4
3rd day:p3=1*1=1

total P=1/9*1/4*1 = 1/(9*4)
this can be done in 2 ways =2/(9*4)=1/18
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by DarkKnight » Wed Feb 18, 2009 6:58 pm
I am getting 14/81.

(9*8*7*1)*2 / (9*2*9*2*9*2)
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by Bidisha_800 » Thu Feb 19, 2009 12:24 am
total number of combination =3*2*3=18

day 1 = (3 X 1 (shoe #1) X3)/18= 9/18

day2 =(2X 1 (shoe#1) X2)/18 = 4/18

day 3 = 1 X 1 X 1/18 =1/18

same can happen with shoe #2

so probability = 2* (9/18)* (4/18)*(1/18)

=1/81
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by Mr2Bits » Thu Feb 19, 2009 8:52 am
https://www.beatthegmat.com/probability- ... html#93981
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by billzhao » Thu Feb 19, 2009 8:43 pm
To solve this, first realize that you don’t need to do calculations for day one; the odds of him wearing any outfit are 1. For day two, he needs to wear one of the two other shirts that he didn’t wear on day one, the same pair of shoes, and one of the two other pants; the probability of this is: 2/3*1/2*2/3=2/9 . For day three, he needs to wear the remaining unworn shirt, the same pair of shoes, and the remaining unworn pants; the probability is:1/3*1/2*1/3=1/18 . So, the probability that he will wear a different shirt and pants on all three days is:2/9*1/18=1/81=(1/3)^4 .
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