Expected value problem... Probability...

Nappy · 12-08-2009, 12:42

ok i dont have the mathlab that enables me to write the symbols correctly but the question is looking for the expected value of the binomial distribution. Ill do my best to make it readable..

P(Y=j) = (n choose j).p(to the power of j). ( 1 -p )(to the power of (n - j))

Would anyone be able to give me a walkthrough of this question. I know the answer is np but I dont understand the solution I have here in front of me.

Thanks alot...

Fremen · 12-08-2009, 13:27

I guess this is the derivation you're using, right?
http://en.wikipedia.org/wiki/Binomia...n_and_variance

The trick is to rewrite $k {{n}\choose{k}}$ in a clever way.
$k {{n}\choose{k}} = k \frac{n!}{k!(n-k)!}$ by definition.

But if we cancel the Ks and factor out an n,
$k \frac{n!}{k!(n-k)!} = n \frac{(n-1)!}{(k-1)!(n-k)!}$

But this is just
$n {{n-1}\choose{k-1}}$

(check this!)

Now if you factor out the n and a P from the sum, what you're left with sums to 1, because it's a sum of binomial probabilities (with parameters p and (n-1))

Post back if you didn't follow, I'll try to clarify.

Nappy · 12-08-2009, 13:33

thats was exactly the problem Thanks a mill, All sorted now!

12-08-2009, 12:42	#1
Nappy Registered User Join Date: Jun 2008 Posts: 135 Adverts \| Friends	Expected value problem... Probability... ok i dont have the mathlab that enables me to write the symbols correctly but the question is looking for the expected value of the binomial distribution. Ill do my best to make it readable.. P(Y=j) = (n choose j).p(to the power of j). ( 1 -p )(to the power of (n - j)) Would anyone be able to give me a walkthrough of this question. I know the answer is np but I dont understand the solution I have here in front of me. Thanks alot...

12-08-2009, 13:27	#2
Fremen Registered User Join Date: May 2006 Location: arrakis Posts: 1,147 Adverts \| Friends	I guess this is the derivation you're using, right? http://en.wikipedia.org/wiki/Binomia...n_and_variance The trick is to rewrite $k {{n}\choose{k}}$ in a clever way. $k {{n}\choose{k}} = k \frac{n!}{k!(n-k)!}$ by definition. But if we cancel the Ks and factor out an n, $k \frac{n!}{k!(n-k)!} = n \frac{(n-1)!}{(k-1)!(n-k)!}$ But this is just $n {{n-1}\choose{k-1}}$ (check this!) Now if you factor out the n and a P from the sum, what you're left with sums to 1, because it's a sum of binomial probabilities (with parameters p and (n-1)) Post back if you didn't follow, I'll try to clarify. Last edited by Fremen; 12-08-2009 at 13:33.

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12-08-2009, 13:33	#3
Nappy Registered User Join Date: Jun 2008 Posts: 135 Adverts \| Friends	thats was exactly the problem Thanks a mill, All sorted now!