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# A man chooses an outfit from 3 different shirts, 2 different

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Senior Manager
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A man chooses an outfit from 3 different shirts, 2 different [#permalink]  09 Sep 2005, 21:37
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A man chooses an outfit from 3 different shirts, 2 different 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 the shoes are the only item of clothing which he wears on all 3 days?

(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
Senior Manager
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Let me know if this is correct.

I get 1) ((1/3)^6)*((1/2)^3)

Total Outcomes: 3C1*2C1*3C1 = 18
Favorable outcome = 1/18 (Only One pair of shoes to pick from 18 outcomes)
So Probability he will wear shoes all three days = (1/18)*(1/18)*(1/18)
18 can be broken into 2^1*3^2
So we get ((1/3)^6)*((1/2)^3) as the final probability.
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-Vikram

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This question is not making sense to me. He selects shoes every day is already mentioned.

Can someone explain a solution if the question is valid?

Also, @Vikramm:
I think total choices for 3 days are 3^3 * 2^3 * 3^3. However, question is not clear to me to find favorable choices.

Thanks.
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1 min & A. There are three citerion

- One pair of shoes for 3 day = 1/2 x 1/2 x 1/2
- Different shirt each day = 1/3 x 1/3 x 1/3
- Different pant each day = 1/3 x 1/3 x 1/3

---> (1/3)^6 x (1/2)^3
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Senior Manager
Joined: 30 Oct 2004
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dil66 wrote:
This question is not making sense to me. He selects shoes every day is already mentioned.

Can someone explain a solution if the question is valid?

Also, @Vikramm:
I think total choices for 3 days are 3^3 * 2^3 * 3^3. However, question is not clear to me to find favorable choices.

Thanks.

You are right. But I think the question asks... what's the probability that he "wears" only shoes.
Hence, I constructed the outcomes based on the selection criteria given... and then the probability that only shoes are worn. I should have mentioned "Total Outcomes based on Selection criteria" in my response.
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Total events: 3C1 * 2C1 * 3C1 = 18

Day 1, P (he selected only one shoes): 1/18

Day 2, P (he selected only one shoes): 1/18

Day 3, P (he selected only one shoes): 1/18

Answer: 1/18 * 1/18 * 1/18 which is A.

Tough one...
Intern
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i think it should be (1/3)^4.

1st day: 1
2nd day : 2/3 * 1/2 * 2/3 = 2/9
3rd day : 1/3 * 1/2 * 1/3 = 1/18

Prob of same shoes but diff shirt & diff pant for 3 days = 1*2/9*1/18 = 1/81

Lemme know if i am missing anything here!
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