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September 20, 2025, 02:23:08 pm

Author Topic: Challenging Binomial?  (Read 537 times)  Share 

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kenhung123

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Challenging Binomial?
« on: September 05, 2010, 10:47:18 pm »
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OK this question is doing my head in. There is a total of 30 people which contains 10 who supports, does this simply imply that the success probability=1/3?
I tried n=11, p=1/3 and x=4 for binomialpdf, didn't seem to get the right answer. Is it not as I thought it was done?

f.sharp

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Re: Challenging Binomial?
« Reply #1 on: September 05, 2010, 10:52:03 pm »
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0.238?

kenhung123

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Re: Challenging Binomial?
« Reply #2 on: September 05, 2010, 10:54:45 pm »
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Yea, thats what I got. But the answer is 0.298 unless its wrong?

Yitzi_K

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Re: Challenging Binomial?
« Reply #3 on: September 05, 2010, 10:57:29 pm »
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I think this is a hypergeometric distribution, which is no longer on the course.
« Last Edit: September 05, 2010, 11:05:52 pm by Yitzi_K »
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kenhung123

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Re: Challenging Binomial?
« Reply #4 on: September 05, 2010, 11:03:44 pm »
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Thank goodness.

UncleXxx

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Re: Challenging Binomial?
« Reply #5 on: September 05, 2010, 11:19:25 pm »
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 (10/30) x (9/29) x (8/28) x (7/27) x (20/26) x (19/25) x (18/24) x (17/23) x (16/22) x (15/21) x (14/20) = (1292/1420715)

(1292/140715) x nCr(11,4) = 0.298005

Don't know any other faster way on how to do this. buh yehr.
« Last Edit: September 05, 2010, 11:21:23 pm by UncleXxx »