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Math Help - Posterior distribution

  1. #1
    Super Member craig's Avatar
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    Posterior distribution

    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
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Multiply both numerator and denominator of what you got by 6^n, and you will get the result you're given...
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by Moo View Post
    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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