1. test

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

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

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

$\displaystyle = \frac{P(A_{i}) \times P(E \mid A_{i})}{P(A_{1}) \times P(E \mid A_{1}) + P(A_{2}) \times P(E \mid A_{2}) + ... + P(A_{n}) \times P(E \mid A_{n})}$

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

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

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

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