Posterior distribution

**craig** · February 24th 2010, 12:23 PM

Hi, I'm pretty sure this question belongs here, but if it's more just algebra please move to a better place

I have $P_{X,Y}(x,y) = \frac{1}{2}\left(^{n}C_{x}\right)\left(\frac{1}{6} \right)^x\left(\frac{5}{6}\right)^{n-x}$

and

I have $P_{X}(x) = \frac{1}{2}\left(^{n}C_{x}\right)\left(\frac{1}{6} \right)^x\left(\frac{5}{6}\right)^{n-x} + \frac{1}{2}\left(^{n}C_{x}\right)\left(\frac{1}{2} \right)^n$

Now the question is asking me for $P_{Y|X}(y|x)$ , using the formula $P_{Y|X}(y|x) = \frac{P_{X,Y}(x,y)}{P_{X}(x)}$ I've managed to get:

$\frac{\left(\frac{5^{n-x}}{6^n}\right)}{\left(\frac{5^{n-x}}{6^n}\right) + \left(\frac{1}{2}\right)^n}$ , yet the answer states that it's:

$\frac{5^{n-x}}{5^{n-x}+3^n}$

Hoping someone can help me see where I've gone wrong?

Thanks in advance

**Moo** · February 24th 2010, 12:28 PM

Hello,

Multiply both numerator and denominator of what you got by $6^n$ , and you will get the result you're given...

**craig** · February 24th 2010, 12:59 PM

Originally Posted by Moo

Hello,

Multiply both numerator and denominator of what you got by $6^n$ , and you will get the result you're given...

How did I not see that!! Thankyou

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