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# 4 elements are selected at random and with replacement from

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GMAT Instructor
Joined: 04 Jul 2006
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4 elements are selected at random and with replacement from [#permalink]

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11 Aug 2006, 07:39
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4 elements are selected at random and with replacement from the set {1/6, 1/4,1/3,1/2}. What is the probability that the sum of these elements will be an integer?

Kudos [?]: 341 [0], given: 0

Senior Manager
Joined: 05 Mar 2006
Posts: 348

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11 Aug 2006, 08:08
OK, tell me what I'm doing wrong.

Number of possibilties: 4*4*4*4

Number of possibilites to sum up to an integer with 4 elements:

1) 4*(1/4) = 1, 1 combination

2) 4*(1/2) = 1, 1 combination

3) 2*(1/6) + 2*(1/3) = 1, 6 combinations (4c2)

therefore total combinations = 8

8/(4*4*4*4) = 1/32

Am I missing out other combinations?

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CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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11 Aug 2006, 08:30
Total cases = 4^4 = 256

Instead of this I will consider {2,3,4,6} and sum should be divisible by 12.

2,2,2,6 = 12-------Cases = 4
3,3,3,3 = 12-------Cases = 1
2,2,4,4 = 12-------Cases = 6
6,6,6,6 = 24-------Cases = 1

Prob = 12/256 = 3/64
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Kudos [?]: 336 [0], given: 0

Senior Manager
Joined: 05 Mar 2006
Posts: 348

Kudos [?]: 13 [0], given: 0

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11 Aug 2006, 08:36
Aha,

Thanks PS, I missed the combination of 3*(1/3) + 1/2

4c3 = 4 Combinations

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Manager
Joined: 05 Jul 2006
Posts: 195

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16 Aug 2006, 17:12
PS can you explain how you arrive to these:

2,2,2,6 = 12-------Cases = 4
3,3,3,3 = 12-------Cases = 1
2,2,4,4 = 12-------Cases = 6
6,6,6,6 = 24-------Cases = 1

Kudos [?]: 77 [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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16 Aug 2006, 20:11
lan583 wrote:
PS can you explain how you arrive to these:

2,2,2,6 = 12-------Cases = 4
3,3,3,3 = 12-------Cases = 1
2,2,4,4 = 12-------Cases = 6
6,6,6,6 = 24-------Cases = 1

2,2,2,6
there are four cases
6,2,2,2
2,6,2,2
2,2,6,2
2,2,2,6
This can be calculated using 4!/3! = 4

Similarly for 3,3,3,3 there is only one case.

For 2,2,4,4 there are 6 cases
2,2,4,4
2,4,4,2
2,4,2,4
4,2,2,4
4,2,4,2
4,2,2,2
this can be calculated using 4!/(2!*2!) = 6

For 6,6,6,6 there is only one case.

Hope this helps.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Kudos [?]: 336 [0], given: 0

16 Aug 2006, 20:11
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