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

First, calculate \(P(A)\):

The possible outcomes for the sum of two dice being less than 6 are: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). There are 10 outcomes.

Thus, \(P(A) = \frac{10}{36} = \frac{5}{18} \approx 0.278\).

Next, calculate \(P(B)\):

The possible outcomes for the difference between the numbers being at most 2 are: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (3,5), (4,2), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6). There are 24 outcomes.

Thus, \(P(B) = \frac{24}{36} = \frac{2}{3} \approx 0.667\).

Now, calculate \(P(A \cap B)\):

The outcomes that satisfy both conditions are: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1), (3,2), (4,1). There are 8 outcomes.

Thus, \(P(A \cap B) = \frac{8}{36} = \frac{2}{9} \approx 0.222\).

Check independence:

\(P(A) \times P(B) = \frac{5}{18} \times \frac{2}{3} = \frac{5}{27} \approx 0.185\).

Since \(P(A \cap B) \neq P(A) \times P(B)\), the events are not independent.

To determine if events X and Y are independent, we need to check if \(P(X \cap Y) = P(X) \times P(Y)\).

First, calculate \(P(X)\):

The possible outcomes for the total score to be 4 are (1,3), (2,2), and (3,1). Thus, there are 3 favorable outcomes.

Since there are 36 possible outcomes when a die is thrown twice, \(P(X) = \frac{3}{36}\).

Next, calculate \(P(Y)\):

The first score can be 2 or 5. For each of these, there are 6 possible outcomes for the second die. Thus, there are 12 favorable outcomes.

So, \(P(Y) = \frac{12}{36}\).

Now, calculate \(P(X \cap Y)\):

The outcomes that satisfy both X and Y are (2,2). Thus, there is 1 favorable outcome.

So, \(P(X \cap Y) = \frac{1}{36}\).

Check for independence:

\(P(X) \times P(Y) = \frac{3}{36} \times \frac{12}{36} = \frac{1}{36}\).

Since \(P(X \cap Y) = P(X) \times P(Y)\), events X and Y are independent.

(i) The total number of students studying Music is \(40 + 12 = 52\). The total number of students is 160. Therefore, the probability that a randomly chosen student is studying Music is \(\frac{52}{160} = \frac{13}{40}\) or 0.325.

(ii) To determine independence, we check if \(P(\text{boy}) \times P(\text{Music}) = P(\text{boy and Music})\).

\(P(\text{boy}) = \frac{96}{160}\) since there are \(24 + 40 + 32 = 96\) boys.

\(P(\text{Music}) = \frac{52}{160}\) as calculated earlier.

\(P(\text{boy and Music}) = \frac{40}{160}\) since there are 40 boys studying Music.

Now, \(\frac{96}{160} \times \frac{52}{160} \neq \frac{40}{160}\), so the events are not independent.

(i) Let \(M\) be the event that a student is male, \(F\) be the event that a student is female, and \(H\) be the event that a student likes their curry hot.

We have \(P(M) = \frac{3}{4}\) and \(P(F) = \frac{1}{4}\).

The probability that a male student likes their curry hot is \(P(H|M) = \frac{3}{5}\), so \(P(M \cap H) = P(M) \cdot P(H|M) = \frac{3}{4} \cdot \frac{3}{5} = \frac{9}{20}\).

The probability that a female student likes their curry hot is \(P(H|F) = \frac{4}{5}\), so \(P(F \cap H) = P(F) \cdot P(H|F) = \frac{1}{4} \cdot \frac{4}{5} = \frac{1}{5} = \frac{4}{20}\).

The probability that a student likes their curry hot is \(P(H) = P(M \cap H) + P(F \cap H) = \frac{9}{20} + \frac{4}{20} = \frac{13}{20}\).

The probability that the student is either female, or likes their curry hot, or both is \(P(F \cup H) = P(F) + P(H) - P(F \cap H) = \frac{1}{4} + \frac{13}{20} - \frac{4}{20} = \frac{5}{20} + \frac{13}{20} - \frac{4}{20} = \frac{14}{20} = \frac{7}{10}\).

(ii) To determine independence, check if \(P(M \cap H) = P(M) \cdot P(H)\).

We have \(P(M \cap H) = \frac{9}{20}\) and \(P(M) \cdot P(H) = \frac{3}{4} \cdot \frac{13}{20} = \frac{39}{80}\).

Since \(\frac{9}{20} \neq \frac{39}{80}\), the events are not independent.

(i) The number of silver estate cars is 10. The total number of cars is 160. The probability is \(\frac{10}{160} = \frac{1}{16} = 0.0625\).

(ii) The total number of hatchback cars is \(40 + 26 + 24 = 90\). The probability is \(\frac{90}{160} = \frac{9}{16} = 0.5625\).

(iii) The number of red hatchback cars is 40. The probability of a car being a hatchback is \(\frac{90}{160}\). The probability of a car being red given it is a hatchback is \(\frac{40}{90} = \frac{4}{9}\).

(iv) For independence, \(P(\text{red}) \times P(\text{hatchback}) = \frac{72}{160} \times \frac{90}{160}\) should equal \(\frac{40}{160}\). Calculating, \(\frac{72}{160} \times \frac{90}{160} \neq \frac{40}{160}\), so the events are not independent.