(i) To find the probability that a student prefers swimming, we sum the number of students who prefer swimming and divide by the total number of students.

The total number of students who prefer swimming is \(104 + 31 = 135\).

The probability is \(\frac{135}{400} = 0.3375\).

(ii) To determine if the events are independent, we check if \(P(M \cap S) = P(M) \times P(S)\).

\(P(M) = \frac{180}{400} = 0.45\)

\(P(S) = \frac{135}{400} = 0.3375\)

\(P(M \cap S) = \frac{31}{400} = 0.0775\)

Calculate \(P(M) \times P(S) = 0.45 \times 0.3375 = 0.151875\)

Since \(0.151875 \neq 0.0775\), the events are not independent.

(a) To find \(P(A \cap B)\), we need to determine the outcomes where the score on the red die is divisible by 3 and the sum of the two scores is at least 9.

The possible scores on the red die that are divisible by 3 are 3 and 6.

\(For red die = 3, the possible outcomes for the blue die to make the sum at least 9 are: (3,6).\)

\(For red die = 6, the possible outcomes for the blue die to make the sum at least 9 are: (6,3), (6,4), (6,5), (6,6).\)

Thus, the favorable outcomes are: (3,6), (6,3), (6,4), (6,5), (6,6).

There are 5 favorable outcomes out of 36 possible outcomes, so \(P(A \cap B) = \frac{5}{36}\).

(b) To determine if \(A\) and \(B\) are independent, we check if \(P(A) \times P(B) = P(A \cap B)\).

\(P(A) = \frac{2}{6} = \frac{1}{3}\) since the red die can be 3 or 6.

\(P(B) = \frac{10}{36} = \frac{5}{18}\) since there are 10 outcomes where the sum is at least 9.

\(P(A) \times P(B) = \frac{1}{3} \times \frac{5}{18} = \frac{5}{54}\).

Since \(\frac{5}{54} \neq \frac{5}{36}\), events \(A\) and \(B\) are not independent.

To determine if the events are independent, we need to check if:

\(P(\text{hockey}) \times P(\text{Amos or Benn}) = P(\text{hockey} \cap (\text{Amos or Benn}))\)

Calculate each probability:

\(P(\text{hockey}) = \frac{220}{500} = 0.44\)

\(P(\text{Amos or Benn}) = \frac{86 + 156}{500} = \frac{242}{500} = 0.484\)

\(P(\text{hockey} \cap (\text{Amos or Benn})) = \frac{32 + 72}{500} = \frac{104}{500} = 0.208\)

Check the independence condition:

\(0.44 \times 0.484 = \frac{1331}{6250} \approx 0.213\)

Since \(0.213 \neq 0.208\), the events are not independent.

(i) The number of male students who do not play the piano is the sum of male students who play the guitar and the flute:

\(78 + 42 = 120.\)

The probability that a randomly chosen student is a male who does not play the piano is:

\(\frac{120}{300} = 0.4\).

(ii) To determine if the events are independent, we check if \(P(\text{male}) \times P(\text{not piano}) = P(\text{male} \cap \text{not piano})\).

The probability that a student is male is:

\(P(\text{male}) = \frac{160}{300}\).

The probability that a student does not play the piano is:

\(P(\text{not piano}) = \frac{225}{300}\).

Therefore,

\(P(\text{male}) \times P(\text{not piano}) = \frac{160}{300} \times \frac{225}{300} = \frac{8}{15} \times \frac{3}{4} = \frac{2}{5}\).

The probability that a student is male and does not play the piano is:

\(P(\text{male} \cap \text{not piano}) = \frac{120}{300} = \frac{2}{5}\).

Since \(P(\text{male}) \times P(\text{not piano}) = P(\text{male} \cap \text{not piano})\), the events are independent.

First, calculate the probability of event S (sum is even). The possible sums when two dice are thrown range from 2 to 12. The even sums are 2, 4, 6, 8, 10, and 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 6: (1,5), (2,4), (3,3), (4,2), (5,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 10: (4,6), (5,5), (6,4)
• 12: (6,6)

Total even outcomes = 18. Therefore, \(P(S) = \frac{18}{36} = \frac{1}{2}\).

Next, calculate the probability of event T (sum is less than 6 or a multiple of 4). The sums less than 6 are 2, 3, 4, 5, and the multiples of 4 are 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 3: (1,2), (2,1)
• 4: (1,3), (2,2), (3,1)
• 5: (1,4), (2,3), (3,2), (4,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for T = 16. Therefore, \(P(T) = \frac{16}{36} = \frac{4}{9}\).

Now, calculate \(P(S \cap T)\) (sum is even and either less than 6 or a multiple of 4). The even sums that satisfy T are 2, 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for \(S \cap T\) = 10. Therefore, \(P(S \cap T) = \frac{10}{36} = \frac{5}{18}\).

For events S and T to be independent, \(P(S) \times P(T) = P(S \cap T)\). Calculate \(P(S) \times P(T) = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}\).

Since \(\frac{2}{9} \neq \frac{5}{18}\), the events S and T are not independent.