ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kenhung123 on September 05, 2010, 10:47:18 pm

Title: Challenging Binomial?
Post by: kenhung123 on September 05, 2010, 10:47:18 pm

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?

Title: Re: Challenging Binomial?
Post by: f.sharp on September 05, 2010, 10:52:03 pm

0.238?

Title: Re: Challenging Binomial?
Post by: kenhung123 on September 05, 2010, 10:54:45 pm

Yea, thats what I got. But the answer is 0.298 unless its wrong?

Title: Re: Challenging Binomial?
Post by: Yitzi_K on September 05, 2010, 10:57:29 pm

I think this is a hypergeometric distribution, which is no longer on the course.

Title: Re: Challenging Binomial?
Post by: kenhung123 on September 05, 2010, 11:03:44 pm

Thank goodness.

Title: Re: Challenging Binomial?
Post by: UncleXxx on September 05, 2010, 11:19:25 pm

 (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.