(i) The sum of all probabilities must equal 1:
\(\frac{1}{4} + p + p + \frac{3}{8} + 4p = 1\)
Simplifying, we get:
\(\frac{1}{4} + 2p + \frac{3}{8} + 4p = 1\)
\(\frac{1}{4} + \frac{3}{8} + 6p = 1\)
\(\frac{2}{8} + \frac{3}{8} + 6p = 1\)
\(\frac{5}{8} + 6p = 1\)
\(6p = 1 - \frac{5}{8}\)
\(6p = \frac{3}{8}\)
\(p = \frac{1}{16}\)
(ii) To find \(E(X)\):
\(E(X) = (-1) \times \frac{1}{4} + 0 \times \frac{1}{16} + 1 \times \frac{1}{16} + 2 \times \frac{3}{8} + 4 \times \frac{1}{4}\)
\(E(X) = -\frac{1}{4} + 0 + \frac{1}{16} + \frac{6}{8} + 1\)
\(E(X) = -\frac{1}{4} + \frac{1}{16} + \frac{6}{8} + 1\)
\(E(X) = \frac{-4}{16} + \frac{1}{16} + \frac{12}{16} + \frac{16}{16}\)
\(E(X) = \frac{25}{16}\)
To find \(\text{Var}(X)\):
\(\text{Var}(X) = \sum (x^2 \cdot P(X = x)) - (E(X))^2\)
\(\text{Var}(X) = ((-1)^2 \times \frac{1}{4}) + (0^2 \times \frac{1}{16}) + (1^2 \times \frac{1}{16}) + (2^2 \times \frac{3}{8}) + (4^2 \times \frac{1}{4}) - \left(\frac{25}{16}\right)^2\)
\(\text{Var}(X) = \frac{1}{4} + 0 + \frac{1}{16} + \frac{12}{8} + \frac{16}{4} - \left(\frac{25}{16}\right)^2\)
\(\text{Var}(X) = \frac{1}{4} + \frac{1}{16} + \frac{12}{8} + 4 - \frac{625}{256}\)
\(\text{Var}(X) = \frac{863}{256} \; \text{or} \; 3.37\)
(iii) Given that \(X > 0\), find \(P(X = 2)\):
\(P(X > 0) = P(X = 1) + P(X = 2) + P(X = 4) = \frac{1}{16} + \frac{3}{8} + \frac{1}{4}\)
\(P(X > 0) = \frac{1}{16} + \frac{6}{16} + \frac{4}{16} = \frac{11}{16}\)
\(P(X = 2 | X > 0) = \frac{P(X = 2)}{P(X > 0)} = \frac{\frac{3}{8}}{\frac{11}{16}}\)
\(P(X = 2 | X > 0) = \frac{6}{11} \; \text{or} \; 0.545\)
