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GMAT Instructor
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4 elements are selected at random and with replacement from [#permalink]
 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?
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Senior Manager
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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
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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
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Senior Manager
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Aha,
Thanks PS, I missed the combination of 3*(1/3) + 1/2
4c3 = 4 Combinations
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Manager
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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
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CEO
Joined: 20 Nov 2005
Posts: 2934
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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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
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