Daniel_Feldman
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- Sep 30, 2005
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Solved...Thank you.
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One more probability question
- Thread starter Daniel_Feldman
- Start date
Let \(\displaystyle E_{k}\) be the event that the kth couple is seated together. We want its complement.
\(\displaystyle 1-P\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right)\).
\(\displaystyle \sum_{k=0}^{4}\frac{(-1)^{k}\binom 8 k \cdot 2^{k}\cdot (8-k)!}{8!}\)
Hope I got that right :wink:
EDIT: I just found this. See the bottom of the page, problem #3.
http://pages.cs.wisc.edu/~oufang/discussion/D02.pdf
\(\displaystyle 1-P\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right)\).
\(\displaystyle \sum_{k=0}^{4}\frac{(-1)^{k}\binom 8 k \cdot 2^{k}\cdot (8-k)!}{8!}\)
Hope I got that right :wink:
EDIT: I just found this. See the bottom of the page, problem #3.
http://pages.cs.wisc.edu/~oufang/discussion/D02.pdf
D
Deleted member 4993
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Daniel_Feldman said:Solved...Thank you.
Daniel,
Why did you erase your original question?
Please replace it - so that a student at a future date can get the benifit of teaching that Glactus provided you.