It is currently 17 Oct 2017, 19:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 27 Aug 2005
Posts: 331

Kudos [?]: 194 [0], given: 0

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

### Show Tags

09 Sep 2005, 22:37
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 194 [0], given: 0

Senior Manager
Joined: 30 Oct 2004
Posts: 284

Kudos [?]: 67 [0], given: 0

### Show Tags

11 Sep 2005, 11:05
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.
_________________

-Vikram

Kudos [?]: 67 [0], given: 0

Intern
Joined: 25 Jun 2005
Posts: 23

Kudos [?]: 1 [0], given: 0

Location: Bay Area, CA

### Show Tags

01 Oct 2005, 09:00
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.
_________________

If you can't change the people, change the people.

Kudos [?]: 1 [0], given: 0

Senior Manager
Affiliations: CFA Level 2
Joined: 05 May 2004
Posts: 263

Kudos [?]: 150 [0], given: 0

Location: Hanoi

### Show Tags

01 Oct 2005, 10:13
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
_________________

"Life is like a box of chocolates, you never know what you'r gonna get"

Kudos [?]: 150 [0], given: 0

Senior Manager
Joined: 30 Oct 2004
Posts: 284

Kudos [?]: 67 [0], given: 0

### Show Tags

01 Oct 2005, 13:49
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.
_________________

-Vikram

Kudos [?]: 67 [0], given: 0

Manager
Joined: 25 Aug 2004
Posts: 169

Kudos [?]: 35 [0], given: 0

Location: MONTREAL

### Show Tags

02 Oct 2005, 17:48
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...

Kudos [?]: 35 [0], given: 0

Intern
Joined: 14 Jun 2005
Posts: 37

Kudos [?]: 5 [0], given: 0

### Show Tags

02 Oct 2005, 19:28
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!

Kudos [?]: 5 [0], given: 0

02 Oct 2005, 19:28
Display posts from previous: Sort by