Proofs for the Expectation and Variance (VCE SSCE Mathematical Methods): Quizzes

$a = (1-p)$ and $b = p$

Because $(1-p) + p = 1$, so $1^n = 1$

$E(X) = np$

When $x = 0$, the term equals zero

$z = x - 1$

$\text{Var}(X) = np(1-p)$

$x^2$ is not a factor of $x!$

$n(n-1)p^2$

When $p = 0.5$

$E[X(X-1)] = E(X^2) - E(X)$