(i) To find \(P(X = 2)\), consider the combinations: \((0, 2)\) and \((2, 0)\).

\(P(0, 2) = \frac{6}{10} \times \frac{3}{7}\)

\(P(2, 0) = \frac{3}{10} \times \frac{4}{7}\)

\(P(X = 2) = \frac{6}{10} \times \frac{3}{7} + \frac{3}{10} \times \frac{4}{7} = \frac{30}{70} = \frac{3}{7}\)

(ii) The probability distribution of X is:

(iii) To find \(E(X)\) and \(\text{Var}(X)\):

\(E(X) = \sum x \cdot P(X = x) = \frac{13}{7}\)

\(\text{Var}(X) = \sum x^2 \cdot P(X = x) - (E(X))^2\)

\(\text{Var}(X) = \frac{120}{70} + \frac{208}{70} + \frac{108}{70} - \left(\frac{13}{7}\right)^2 = 2.78\)

(iv) Given \(X = 2\), find \(P(A = 2)\):

\(P(A = 2 \mid X = 2) = \frac{P(A = 2 \cap X = 2)}{P(X = 2)} = \frac{\frac{3}{10} \times \frac{4}{7}}{\frac{30}{70}} = 0.4\)