Calculation of expectation and variance

Revisitng our first example.

Let \(X\) be the number of fours in three tosses of a fair six-sided die. Then
\[ X \sim B \left(3,\frac{1}{6} \right). \]

We would have
\[ P(X = k) = {}^3 C_k \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{3-k}, \quad k = 0,1,2,3. \]

One can verify that
\[ \mathrm{E}(X) = 0 \left(\frac{5}{6}\right)^3 + 1 \left(3 \times \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)\right) + 2 \left(3 \times \left(\frac{5}{6}\right) \left(\frac{1}{6}\right)^2\right) + 3 \left(\frac{1}{6}\right)^3 = \frac{1}{2} = 3 \times \frac{1}{6} = np. \]

Similarly, one can verify that
\[ \begin{align*} \mathrm{E}(X^2) - \left(E(X) \right)^2 & = 0^2 \left(\frac{5}{6}\right)^3 + 1^2 \left(3 \times \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)\right) + 2^2 \left(3 \times \left(\frac{5}{6}\right) \left(\frac{1}{6}\right)^2\right) + 3^2 \left(\frac{1}{6}\right)^3 - \left( \frac{1}{2} \right)^2 \\ & = \frac{5}{12} \\ & = 3 \times \frac{1}{6} \times \frac{5}{6} \\ & = np(1-p). \end{align*} \]