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  1. #1
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    test

    <br />
\left(\begin{array}{cc}n\\n_{1},n_{2}, ..., n_{k}\end{array}\right) = \frac{n!}{n_{1}!n_{2}!...n_{k}!}<br />

    <br />
P(X=k) = \frac{1}{n}<br />

    <br />
E(X) = \frac{n+1}{2}<br />
    <br />
V(X) = \frac{n^{2}-1}{12}<br />

    <br />
= \frac{P(A_{i}) \times P(E \mid A_{i})}{P(A_{1}) \times P(E \mid A_{1}) + P(A_{2}) \times P(E \mid A_{2}) + ... + P(A_{n}) \times P(E \mid A_{n})}<br />
    Last edited by thatloserrsaid; May 25th 2010 at 08:34 AM.
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  2. #2
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    Quote Originally Posted by thatloserrsaid View Post
    <br />
\left(\begin{array}{cc}n\\n_{1},n_{2}, ..., n_{k}\end{array}\right) = \frac{n!}{n_{1}!n_{2}!...n_{k}!}<br />

    <br />
P(X=k) = \frac{1}{n}<br />

    <br />
E(X) = \frac{n+1}{2}<br />
    <br />
V(X) = \frac{n^{2}-1}{12}<br />

    <br />
= \frac{P(A_{i}) \times P(E \mid A_{i})}{P(A_{1}) \times P(E \mid A_{1}) + P(A_{2}) \times P(E \mid A_{2}) + ... + P(A_{n}) \times P(E \mid A_{n})}<br />
    Depending on your preferences you might also like to use \cdot so that P(A_{i}) \times P(E \mid A_{i}) becomes P(A_{i}) \cdot P(E \mid A_{i}).
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