# test

• May 24th 2010, 09:44 AM
thatloserrsaid
test
$
\left(\begin{array}{cc}n\\n_{1},n_{2}, ..., n_{k}\end{array}\right) = \frac{n!}{n_{1}!n_{2}!...n_{k}!}
$

$
P(X=k) = \frac{1}{n}
$

$
E(X) = \frac{n+1}{2}
$

$
V(X) = \frac{n^{2}-1}{12}
$

$
= \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})}
$
• May 30th 2010, 07:11 PM
undefined
Quote:

Originally Posted by thatloserrsaid
$
\left(\begin{array}{cc}n\\n_{1},n_{2}, ..., n_{k}\end{array}\right) = \frac{n!}{n_{1}!n_{2}!...n_{k}!}
$

$
P(X=k) = \frac{1}{n}
$

$
E(X) = \frac{n+1}{2}
$

$
V(X) = \frac{n^{2}-1}{12}
$

$
= \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})}
$

Depending on your preferences you might also like to use \cdot so that $P(A_{i}) \times P(E \mid A_{i})$ becomes $P(A_{i}) \cdot P(E \mid A_{i})$.