(i) To find the probability of event A, we list the pairs where the scores differ by 3 or more: (1,4), (1,5), (1,6), (2,5), (2,6), (3,6) and their reverses: (4,1), (5,1), (6,1), (5,2), (6,2), (6,3). There are 12 such pairs out of 36 possible outcomes.

Thus, \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(ii) For event B, we need the product of the scores to be greater than 8. The pairs are: (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6). There are 20 such pairs.

Thus, \(P(B) = \frac{20}{36} = \frac{5}{9}\).

(iii) Events A and B are not mutually exclusive because \(P(A \cap B) \neq 0\). Specifically, there are 6 outcomes that satisfy both conditions: (2,5), (2,6), (3,6), (5,2), (6,2), (6,3). Therefore, \(P(A \cap B) = \frac{6}{36}\), so they are not mutually exclusive.

(i) To find \(P(M)\), calculate the total number of males: \(206 + 412 = 618\). The total number of people is \(206 + 412 + 358 + 305 = 1281\). Thus, \(P(M) = \frac{618}{1281} \approx 0.482\).

(ii) To find \(P(M \text{ and } E)\), use the number of employed males: \(412\). Therefore, \(P(M \text{ and } E) = \frac{412}{1281} \approx 0.322\).

(iii) To check if \(M\) and \(E\) are independent, calculate \(P(E)\): \(P(E) = \frac{717}{1281}\) (since \(412 + 305 = 717\)). Check if \(P(M) \times P(E) = P(M \text{ and } E)\). \(\frac{618}{1281} \times \frac{717}{1281} \neq \frac{412}{1281}\), so they are not independent.

(iv) Given the person is unemployed, find the probability they are female. Total unemployed is \(206 + 358 = 564\). Thus, \(P(\text{Female} | \text{Unemployed}) = \frac{358}{564} \approx 0.635\).

(a) The tree diagram shows the outcomes of the first and second throws. Each branch represents a possible outcome with a probability of \(\frac{1}{6}\).

(b) To find \(P(A)\), calculate the probability of scoring 5, 6, 7, 8, or 9:

- Score 5: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} = \frac{7}{36}\)

- Score 6: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 7: \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}\)

- Score 8: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

- Score 9: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

Adding these probabilities gives \(P(A) = \frac{12}{36} = \frac{1}{3}\).

(c) To check independence, compare \(P(A \cap B)\) and \(P(A)P(B)\):

- \(P(B) = \frac{1}{3}\), \(P(A \cap B) = \frac{6}{36} = \frac{1}{6}\)

- \(P(A)P(B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\)

Since \(\frac{1}{6} \neq \frac{1}{9}\), events A and B are not independent.

(d) To find \(P(B | A')\), use the formula:

\(P(B | A') = \frac{P(B \cap A')}{P(A')} = \frac{\frac{6}{36}}{\frac{2}{3}} = \frac{1}{4} = 0.25\)

(i) For events A and B to be independent, \(P(A) \times P(B)\) should equal \(P(A \text{ and } B)\). Calculate \(P(A) \times P(B) = 0.3 \times 0.8 = 0.24\). Since \(0.24 \neq 0.4\), events A and B are not independent.

(ii) For events A and B to be mutually exclusive, \(P(A \text{ and } B)\) should be 0. Since \(P(A \text{ and } B) = 0.4 \neq 0\), events A and B are not mutually exclusive.

(a) The total number of students who play the guitar is \(44 + 38 = 82\).

The probability that a student plays the guitar is \(\frac{82}{180} = \frac{41}{90}\).

(c) Let \(F\) be the event that the student is female, and \(G\) be the event that the student plays the guitar.

\(P(F) = \frac{100}{180} = \frac{5}{9} \approx 0.5556\)

\(P(G) = \frac{82}{180} = \frac{41}{90} \approx 0.4556\)

\(P(F \cap G) = \frac{38}{180} = \frac{19}{90} \approx 0.2111\)

To check for independence, we compare \(P(F \cap G)\) with \(P(F) \times P(G)\).

\(P(F) \times P(G) = \frac{5}{9} \times \frac{41}{90} = \frac{205}{810} = \frac{41}{162} \approx 0.2531\)

Since \(P(F \cap G) \neq P(F) \times P(G)\), the events are not independent.