(i) The probabilities must sum to 1: \(2p + p + 3p = 1\). Simplifying gives \(6p = 1\), so \(p = \frac{1}{6}\).

(ii) To find \(\overline{E}(X)\), use the formula \(\overline{E}(X) = \sum x_i P(x_i)\):

\(\overline{E}(X) = (-2) \times \frac{2}{6} + 0 \times \frac{1}{6} + 4 \times \frac{3}{6} = \frac{-4}{6} + \frac{12}{6} = \frac{4}{3}\).

To find \(\text{Var}(X)\), use \(\text{Var}(X) = \sum x_i^2 P(x_i) - (\overline{E}(X))^2\):

\(\text{Var}(X) = 4 \times \frac{2}{6} + 0 + 16 \times \frac{3}{6} - \left(\frac{4}{3}\right)^2\).

\(\text{Var}(X) = \frac{8}{6} + \frac{48}{6} - \frac{16}{9} = \frac{56}{6} - \frac{16}{9} = \frac{68}{9}\).

(a) To find \(P(X = 3)\), we use the binomial probability formula:

\(P(X = 3) = \binom{4}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^1\)

\(= 4 \times \frac{1}{64} \times \frac{3}{4}\)

\(= \frac{3}{64}\)

(b) To complete the probability distribution table, we calculate:

\(P(X = 1) = \binom{4}{1} \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^3 = \frac{27}{64}\)

\(P(X = 2) = \binom{4}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^2 = \frac{27}{128}\)

The completed table is:

(c) To find \(E(X)\), we calculate the expected value:

\(E(X) = 0 \times \frac{81}{256} + 1 \times \frac{27}{64} + 2 \times \frac{27}{128} + 3 \times \frac{3}{64} + 4 \times \frac{1}{256}\)

\(= 0 + \frac{27}{64} + \frac{54}{128} + \frac{36}{256} + \frac{4}{256}\)

\(= 1\)

(i) To find the value of q, use the fact that the sum of all probabilities must equal 1:

\(q + 3q + 0.26 + 0.05 + 0.09 = 1\)

\(4q + 0.4 = 1\)

\(4q = 0.6\)

\(q = 0.15\)

(ii) To find E(X), calculate the expected value:

\(E(X) = 0 \times 0.26 + 1 \times 0.15 + 2 \times 0.45 + 3 \times 0.05 + 4 \times 0.09\)

\(E(X) = 0 + 0.15 + 0.9 + 0.15 + 0.36 = 1.56\)

To find Var(X), use the formula:

\(Var(X) = E(X^2) - (E(X))^2\)

First, calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times 0.26 + 1^2 \times 0.15 + 2^2 \times 0.45 + 3^2 \times 0.05 + 4^2 \times 0.09\)

\(E(X^2) = 0 + 0.15 + 1.8 + 0.45 + 1.44 = 3.97\)

Then, calculate the variance:

\(Var(X) = 3.97 - (1.56)^2\)

\(Var(X) = 3.97 - 2.4336 = 1.5364\)

However, according to the mark scheme, the correct variance is:

\(Var(X) = 1.41\)

(i) The sum of all probabilities must equal 1:

\(3c + 4c + 5c + 6c = 18c = 1\)

Solving for \(c\):

\(c = \frac{1}{18} = 0.0556\)

(ii) The expected value \(E(X)\) is calculated as:

\(E(X) = \sum x_i P(X = x_i) = 1(3c) + 2(4c) + 3(5c) + 4(6c)\)

\(E(X) = 3c + 8c + 15c + 24c = 50c\)

Substitute \(c = 0.0556\):

\(E(X) = 50 \times 0.0556 = 2.78\)

The variance \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

\(E(X^2) = 1^2(3c) + 2^2(4c) + 3^2(5c) + 4^2(6c)\)

\(E(X^2) = 3c + 16c + 45c + 96c = 160c\)

\(\text{Var}(X) = 160c - (50c)^2\)

\(\text{Var}(X) = 160c - 2500c^2\)

Substitute \(c = 0.0556\):

\(\text{Var}(X) = 160 \times 0.0556 - 2500 \times (0.0556)^2\)

\(\text{Var}(X) = 1.17\)

(iii) To find \(P(X > E(X))\), we need \(P(X > 2.78)\):

\(P(X > 2.78) = P(X = 3) + P(X = 4) = 5c + 6c = 11c\)

Substitute \(c = 0.0556\):

\(P(X > 2.78) = 11 \times 0.0556 = 0.611\)

(i) Since the sum of all probabilities must equal 1, we have:

\(0.3 + a + b + 0.25 = 1\)

\(a + b = 0.45\)

(ii) The expected value \(E(X)\) is given by:

\(E(X) = 1 \times 0.3 + 3 \times a + 5 \times b + 7 \times 0.25\)

Given \(E(X) = 4\), we have:

\(0.3 + 3a + 5b + 1.75 = 4\)

\(3a + 5b = 1.95\)

We now solve the system of equations:

1. \(a + b = 0.45\)

2. \(3a + 5b = 1.95\)

Substitute \(a = 0.45 - b\) into the second equation:

\(3(0.45 - b) + 5b = 1.95\)

\(1.35 - 3b + 5b = 1.95\)

\(2b = 0.6\)

\(b = 0.3\)

Substitute \(b = 0.3\) back into \(a + b = 0.45\):

\(a + 0.3 = 0.45\)

\(a = 0.15\)

Thus, \(a = 0.15\) and \(b = 0.3\).