(i) To find the probability of obtaining exactly 1 head and 2 tails, consider the scenarios: HTT, THT, and TTH.

For HTT: \(P(H) = \frac{2}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(HTT) = \frac{2}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For THT: \(P(T) = \frac{1}{3}, P(H) = \frac{2}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(THT) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For TTH: \(P(T) = \frac{1}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(H) = \frac{1}{4}\) for Coin B.

\(P(TTH) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\).

Total probability: \(\frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{12}{36} + \frac{1}{36} = \frac{13}{36}\).

(ii) The probability distribution table is:

(iii) The expectation \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{3}{36} + 1 \times \frac{13}{36} + 2 \times \frac{16}{36} + 3 \times \frac{4}{36}\).

\(E(X) = \frac{0}{36} + \frac{13}{36} + \frac{32}{36} + \frac{12}{36} = \frac{57}{36} = \frac{19}{12} \approx 1.58\).

(i) The probability of choosing exactly 2 paperback books is calculated using combinations. The number of ways to choose 2 paperback books from 6 is \(\binom{6}{2}\), and the number of ways to choose 2 hardback books from 2 is \(\binom{2}{2}\). The total number of ways to choose 4 books from 8 is \(\binom{8}{4}\).

Thus, the probability is:

\(P(\text{exactly 2}) = \frac{\binom{6}{2} \cdot \binom{2}{2}}{\binom{8}{4}} = \frac{15}{70} = \frac{3}{14}\)

(ii) The probability distribution table for X is:

(iii) Given \(E(X) = 3\), we find \(\text{Var}(X)\) using the formula \(\text{Var}(X) = \sum x^2 p - (E(X))^2\).

\(\text{Var}(X) = \frac{12}{14} + \frac{72}{14} + \frac{48}{14} - 3^2\)

\(= \frac{3}{7}\)

To find the probability distribution of \(X\), we consider the possible values of \(X\): 0, 1, or 2.

Case 1: \(X = 0\)

This means no 5s are chosen. There are 6 non-5 digits (1s and 3s) and 3 5s. The probability of choosing two non-5s is:

\(P(0) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12}\)

Case 2: \(X = 1\)

This means one 5 is chosen. From part (ii), \(P(1) = 0.5\).

Case 3: \(X = 2\)

This means two 5s are chosen. From part (i), \(P(2) = \frac{6}{72} = \frac{1}{12}\).

The probability distribution table is:

(i) To find the probability that the sum of the numbers on the three cards is 11, consider the possible combinations: (3, 4, 4), (4, 3, 4), or (4, 4, 3). The probability for each combination is calculated as follows:

\(\text{Prob} = \left( \frac{4}{10} \times \frac{6}{9} \times \frac{5}{8} \right) \times 3C1 = \frac{360}{720} = \frac{1}{2}\)

Alternatively, using combinations:

\(\frac{\binom{6}{2} \times \binom{4}{1}}{\binom{10}{3}} = \frac{1}{2}\)

(ii) The probability distribution table for the sum of the numbers on the three cards is:

Calculations for each sum:

• \(P(3, 3, 3) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720}\)
• \(P(3, 3, 4) = \frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} \times 3C1 = \frac{216}{720}\)
• \(P(4, 4, 4) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720}\)

(i) If the coin shows a head, the smallest sum of two dice throws is 2 (1+1). Therefore, X = 1 can only occur if the coin shows a tail and the first die shows 1. The probability of a tail is \(\frac{1}{2}\) and the probability of rolling a 1 is \(\frac{1}{4}\). Thus, \(P(X = 1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

(ii) For X = 3, if the coin shows a head, possible outcomes are (1,2) and (2,1), each with probability \(\frac{1}{16}\). If the coin shows a tail, the first die must show 3, with probability \(\frac{1}{8}\). Therefore, \(P(X = 3) = \frac{2}{32} + \frac{1}{8} = \frac{3}{16}\).

(iii) Completing the table:

(iv) Events Q and R are exclusive because if a tail is thrown, X cannot be 7. Thus, \(P(Q \cap R) = 0\).