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Percentages Question
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.Roy
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O.P. RE: Percentages Question
Yeah Thanks So much, I thought something was wrong.

Thanks All
07-02-2006 06:55 PM
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markee
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RE: Percentages Question
quote:
Originally posted by Grue
quote:
Originally posted by haydn
The probabilty of getting at least 2 should be Probability of getting two (P2) + P3 + P4 + P5 (As said above).

A simpler way to determine it would be to work out the probability of one being picked plus the probability of none being picked and subtract this amount from one. I think markee's calculation was off after he didn't include the possibility of none being picked.

That is exactly right so working it out....
quote:
Originally posted by markee

So there are 5 possibilities (for the 1 out of 5 trails) ABBBB BABBB BBABB BBBAB BBBBA

1/10*9/10*9/10*9/10*9/10*5 = 32.805% of it NOT being at least 2 out of 5


And the probability of of it being 0 out of 5 would be 9/10*9/10*9/10*9/10*9/10 = .59049 or 59.049%
So 100%-59.049%-32.805%=8.146% which is less than 10% :)

Thanks for fixing that, I'm used to using nPr and nCr for things so I forget a bit of the binomial stuff, sorry about that .Roy.
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07-03-2006 05:44 AM
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