(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 \(P(A = 2)\):

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

(i) The possible values of \(X\) are 0, 1, and 2. The probability distribution is:

(ii) To find \(E(X)\), use the formula:

\(E(X) = \sum x P(X = x) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = \frac{8}{7}\).

To find \(\text{Var}(X)\), use the formula:

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

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

\(E(X^2) = 0^2 \times \frac{1}{7} + 1^2 \times \frac{4}{7} + 2^2 \times \frac{2}{7} = \frac{12}{7}\).

Then, \(\text{Var}(X) = \frac{12}{7} - \left(\frac{8}{7}\right)^2 = \frac{12}{7} - \frac{64}{49} = \frac{20}{49}\).

(i) To find \(P(X = 40)\), we need to find the value of \(r\) such that \(\frac{120}{r} = 40\). Solving for \(r\), we get:

\(40 = \frac{120}{r}\)

\(r = \frac{120}{40} = 3\)

Now, substitute \(r = 3\) into the probability formula:

\(P(X = 40) = \frac{3}{45} = \frac{1}{15}\)

(ii) Construct the probability distribution table for \(X\):

(iii) The modal value of \(X\) is the value with the highest probability. From the table, \(X = 40\) has the highest probability of \(\frac{3}{45}\).

(iv) To find the probability that \(X\) lies between 18 and 100, sum the probabilities for \(X = 20, 24, 30, 40, 60\):

\(P(18 < X < 100) = \frac{6}{45} + \frac{5}{45} + \frac{4}{45} + \frac{3}{45} + \frac{2}{45}\)

\(= \frac{20}{45} = \frac{4}{9}\)

(ii) To find the probability distribution of \(X\), count the occurrences of each sum in the table and divide by the total number of outcomes (36). The probabilities are:

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

\(E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + 6 \times \frac{5}{36} + 7 \times \frac{4}{36} + 8 \times \frac{4}{36} + 9 \times \frac{4}{36} + 10 \times \frac{4}{36} + 11 \times \frac{8}{36} + 12 \times \frac{4}{36}\)

\(= \frac{312}{36} = \frac{26}{3} \approx 8.67\)

(iv) The probability that \(X\) is greater than \(E(X)\) is:

\(P(X > 8.67) = P(X = 9, 10, 11, 12) = \frac{4}{36} + \frac{4}{36} + \frac{8}{36} + \frac{4}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556\)

(i) The tree diagram should have branches for each call attempt with probabilities 0.5 for Answered (A) and 0.5 for Unanswered (U). There should be 4 levels of branches, representing up to 4 attempts.

(ii) The probability distribution table is completed as follows:

(iii) The expected number of unanswered calls \(E(X)\) is calculated as:

\(E(X) = \sum x \cdot P(X = x) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} + 4 \cdot \frac{1}{16}\)

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

\(E(X) = \frac{1}{4} + \frac{1}{4} + \frac{3}{16} + \frac{4}{16}\)

\(E(X) = \frac{1}{2} + \frac{7}{16} = \frac{15}{16}\)

Thus, \(E(X) = \frac{15}{16}\) or approximately 0.938 or 0.9375.