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PR Probability Problem, Pls Help

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PR Probability Problem, Pls Help

by mnjoosub » Thu Mar 20, 2008 6:12 am
The question is:

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
:?: :x
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by madhavi » Thu Mar 20, 2008 8:42 am
On the first day, man has to wear 1 shirt from 3 shirts, 1 pant from 3 pants and 1 shoe from 2 shoes

3C1/3*3C1/3*2C1/2

On the second day, man has to wear 1 shirt of the remaining 2 shirts, 1 pant from the remaining 2 pants and the same shoe

2C1/3*2C1/3*1C1/2

On the third day, man has to wear 1 shirt of the remaining 1 shirt, 1 pant from the remaining 1 pant and the same shoe

1C1/3*1C1/3*1C1/2

So probability for all 3 days is:

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

Hope that helps!
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by mnjoosub » Thu Mar 20, 2008 8:58 am
That's great, the OA = C.

But Y you keep dividing by 3 and 2 on the second and third day? :?
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by madhavi » Thu Mar 20, 2008 9:08 am
No of ways of selecting a pant/shirt as per the case
__________________________________________

The total no of possibilities


The denomitor is always the no of options available that is 2 if case is shoes else 3. The numerator varies depnding on the case, the particular selection.

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by madhavi » Thu Mar 20, 2008 9:13 am
possible outcomes in favour of the case/ total outcomes
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by mnjoosub » Thu Mar 20, 2008 9:28 am
Well Madhavi,

On the second day, as the shirt and the pant is not re-used, the possible outcomes reduces to 2 for the pants and shirts but remains 2 for the shoes and on the third day reduces to 1. Isn't it :?:

Please Help, I am a bit rusted on Prob. and Combinatorics.
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by madhavi » Thu Mar 20, 2008 9:35 am
Lets think of a six faced dice.

Possibility of getting a 2. How do you handle this?

You work out the options for this in the numerator and the denominator has all the possible outcomes. Right?

Similarly, on the second day you have 3 shirts to select from. But you want to check for the event where you would not want to repeat a shirt.

Hence, 3 in the denominator
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by mnjoosub » Thu Mar 20, 2008 9:50 am
IC, Many thanks to u Madhavi.

I think that I need some more of these kind of questions to work on in order to re-master my high school Maths :lol: :D

Thanks
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by madhavi » Thu Mar 20, 2008 10:06 am
GMAT Math is always trricky with counting and geometry. A thorough round of revision is a must. Good luck and keep going.

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