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Probability

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Junior | Next Rank: 30 Posts
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by raptor1985 » Mon Mar 12, 2012 2:26 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)

OA : C
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by Jim@StratusPrep » Tue Mar 13, 2012 7:40 pm
He needs to wear 3 different shirts:
(3/3)(2/3)(1/3)= 2/9

He needs to wear 3 different pants:
(3/3)(2/3)(1/3)= 2/9

And, the same shoe for 3 days:
(2/2)(1/2)(1/2)= 1/4

Multiply these altogether:
2/9(2/9)(1/4)= 1/81 or C
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by icemanKK » Wed Mar 14, 2012 6:32 am
Hi,

Can the above problem be split as follows. I do know that the final answer achieved is wrong as the probability is greater than 1 (which cannot happen). So would need help ....

Day 1 :

Probability of choosing a pant : 1/3
Probability of choosing a shirt : 1/3
Probability of choosing a shoe : 1/2

Day 2 :
Since the set of clothes need to be different from Day 1 the choices reduce to 2 shirts, 2 pants and
1 pair of shoe (the same that was chosen on day 1)

Therefore :
Probability of choosing a pant : 1/2
Probability of choosing a shirt : 1/2
Probability of choosing a shoe : 1


Day 3 :
Since the set of clothes need to be different from Day 1 and Day 2 the choices reduce to 1 shirt, 1 pant and 1 pair of shoe (the same that was chosen on day 1)

Therefore :
Probability of choosing a pant : 1
Probability of choosing a shirt : 1
Probability of choosing a shoe : 1

So Total Probability is :

Total Probability is (1/3)*(1/2)*(1/3) + (1/2)*(1/2) + 1 which is > 1.

Need some help on this. Can someone elaborate on what is wrong.
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