(a) To find the probability of obtaining exactly one head, consider two scenarios: the biased coin shows a head and the three fair coins show tails, or the biased coin shows a tail and one of the fair coins shows a head.

Scenario 1: Biased coin shows head, three fair coins show tails:

\(Probability = 0.6 × (0.5)^3 = 0.6 × 0.125 = 0.075\)

Scenario 2: Biased coin shows tail, one fair coin shows head:

\(Probability = 0.4 × (0.5)^3 × 3 = 0.4 × 0.125 × 3 = 0.15\)

\(Total probability = 0.075 + 0.15 = 0.225\)

(b) To complete the probability distribution table, use the given probabilities and the fact that the sum of probabilities must equal 1.

\((c) Given E(X) = 2.1, use the variance formula:\)

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

Calculate E(X^2) using the probabilities:

\(E(X^2) = 0^2 × 0.05 + 1^2 × 0.225 + 2^2 × 0.375 + 3^2 × 0.275 + 4^2 × 0.075\)

\(= 0 + 0.225 + 1.5 + 2.475 + 1.2 = 5.4\)

\(Var(X) = 5.4 - (2.1)^2 = 5.4 - 4.41 = 0.99\)

First, use the fact that the sum of probabilities must equal 1:

\(0.12 + p + q + 0.16 + 0.3 = 1\)

Simplifying gives:

\(p + q = 0.42\)

Next, use the given expected value \(E(X) = 0.28\):

\(-2(0.12) - 1(p) + 0.5(q) + 1(0.16) + 2(0.3) = 0.28\)

This simplifies to:

\(-0.24 - p + 0.5q + 0.16 + 0.6 = 0.28\)

\(-p + 0.5q = -0.24\)

Now solve the system of equations:

1. \(p + q = 0.42\)
2. \(-p + 0.5q = -0.24\)

Add the two equations to eliminate \(p\):

\((p + q) + (-p + 0.5q) = 0.42 - 0.24\)

\(1.5q = 0.18\)

\(q = 0.12\)

Substitute \(q = 0.12\) back into \(p + q = 0.42\):

\(p + 0.12 = 0.42\)

\(p = 0.3\)

(a) The sum of probabilities must equal 1:

\(p + q + 0.6 + 0.05 = 1\)

\(p + q + 0.65 = 1\)

\(p + q = 0.35\) (1)

Using the expected value \(E(X) = 0.55\):

\(0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\)

\(p + 2q + 0.15 = 0.55\)

\(p + 2q = 0.4\) (2)

Solving equations (1) and (2):

From (1): \(p = 0.35 - q\)

Substitute into (2):

\(0.35 - q + 2q = 0.4\)

\(0.35 + q = 0.4\)

\(q = 0.05\)

Substitute \(q = 0.05\) into (1):

\(p + 0.05 = 0.35\)

\(p = 0.3\)

(b) The variance \(\text{Var}(X)\) is given by:

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

Calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times 0.6 + 1^2 \times 0.3 + 2^2 \times 0.05 + 3^2 \times 0.05\)

\(E(X^2) = 0 + 0.3 + 0.2 + 0.45\)

\(E(X^2) = 0.95\)

Thus, \(\text{Var}(X) = 0.95 - 0.55^2\)

\(\text{Var}(X) = 0.95 - 0.3025\)

\(\text{Var}(X) = 0.6475\)

The sum of all probabilities must equal 1:

\(p + p + 0.1 + q + q = 1\)

Simplifying, we get:

\(2p + 0.1 + 2q = 1\)

Given that \(P(X \geq 0) = 3P(X < 0)\), we have:

\(0.1 + 2q = 3(2p)\)

Now, solve the system of equations:

1. \(2p + 0.1 + 2q = 1\)
2. \(0.1 + 2q = 6p\)

From equation (2), express \(q\) in terms of \(p\):

\(2q = 6p - 0.1\)

\(q = 3p - 0.05\)

Substitute \(q = 3p - 0.05\) into equation (1):

\(2p + 0.1 + 2(3p - 0.05) = 1\)

\(2p + 0.1 + 6p - 0.1 = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Substitute \(p = \frac{1}{8}\) back to find \(q\):

\(q = 3 \times \frac{1}{8} - 0.05\)

\(q = \frac{3}{8} - \frac{1}{20}\)

\(q = \frac{15}{40} - \frac{2}{40}\)

\(q = \frac{13}{40}\)

(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 \(P(X = 2)\):

\(P(X > 0) = P(X = 1) + P(X = 2) + P(X = 4) = \frac{1}{16} + \frac{3}{8} + \frac{1}{4}\)

\(P(X > 0) = \frac{1}{16} + \frac{6}{16} + \frac{4}{16} = \frac{11}{16}\)

\(P(X = 2 | X > 0) = \frac{P(X = 2)}{P(X > 0)} = \frac{\frac{3}{8}}{\frac{11}{16}}\)

\(P(X = 2 | X > 0) = \frac{6}{11} \; \text{or} \; 0.545\)