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.

First, calculate the probability of event R (product is 12). The possible pairs are (3, 4), (4, 3), (2, 6), and (6, 2). Thus,

\(P(R) = \frac{4}{36} = \frac{1}{9}\).

Next, calculate the probability of event T (one number is odd, one is even). There are 18 such pairs out of 36 total possibilities, so

\(P(T) = \frac{18}{36} = \frac{1}{2}\).

Now, calculate the probability of both events occurring together, \(P(R \cap T)\). The pairs (3, 4) and (4, 3) satisfy both conditions, so

\(P(R \cap T) = \frac{2}{36} = \frac{1}{18}\).

Check for independence by verifying if \(P(R) \times P(T) = P(R \cap T)\):

\(\frac{1}{9} \times \frac{1}{2} = \frac{1}{18}\).

Since \(P(R) \times P(T) = P(R \cap T)\), the events are independent.

(i) The number of red or silver cars manufactured in Korea is \(34 + 30 = 64\). The probability is \(\frac{64}{250} = 0.256\).

(ii) The number of cars not manufactured in Japan is \(250 - (26 + 22 + 12) = 190\). The probability is \(\frac{190}{250} = 0.76\).

(iii) Calculate \(P(X)\), \(P(Y)\), and \(P(X \cap Y)\):

\(P(X) = \frac{80}{250} = \frac{8}{25}\)

\(P(Y) = \frac{100}{250} = \frac{2}{5}\)

\(P(X \cap Y) = \frac{32}{250} = \frac{16}{125}\)

Check independence: \(P(X) \times P(Y) = \frac{8}{25} \times \frac{2}{5} = \frac{16}{125}\)

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

(i) To complete the table, we use the given information:

• Total number of adults = 30
• Right-handed adults = 25
• Adults who wear spectacles = 8
• Right-handed and do not wear spectacles = 19

From this, we can fill in the table:

• Right-handed and wears spectacles = 25 - 19 = 6
• Not right-handed = 30 - 25 = 5
• Not right-handed and wears spectacles = 8 - 6 = 2
• Not right-handed and does not wear spectacles = 5 - 2 = 3

The completed table is:

(ii) To determine if events X and Y are independent, we check if:

\(P(X \cap Y) = P(X) \times P(Y)\)

Calculate the probabilities:

• \(P(X) = \frac{25}{30}\)
• \(P(Y) = \frac{8}{30}\)
• \(P(X \cap Y) = \frac{6}{30}\)

Check independence:

\(P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}\)

\(P(X \cap Y) = \frac{6}{30} = \frac{1}{5}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.

(i) To find the probability of event R, we list the pairs where one score is exactly 3 greater than the other: \((1, 4), (2, 5), (3, 6), (4, 7), (5, 8)\). Each pair can occur in two ways (e.g., \((1, 4)\) and \((4, 1)\)). Thus, there are \(5 \times 2 = 10\) favorable outcomes. The total number of outcomes when the die is thrown twice is \(8 \times 8 = 64\). Therefore, \(P(R) = \frac{10}{64}\).

(ii) For event S, we need the product of the scores to be more than 19. The pairs satisfying this condition are \((3, 8), (3, 7), (4, 8), (4, 7), (4, 6), (4, 5), (5, 8), (5, 7), (5, 6), (6, 8), (6, 7), (7, 8)\). Additionally, \((5, 5), (6, 6), (7, 7), (8, 8)\) also satisfy the condition. The total number of favorable outcomes is \(24 + 4 = 28\). Thus, \(P(S) = \frac{28}{64}\).

(iii) To determine if R and S are independent, we check if \(P(R \cap S) = P(R) \times P(S)\). From the mark scheme, \(P(R \cap S) = \frac{4}{64}\). Calculating \(P(R) \times P(S) = \frac{10}{64} \times \frac{28}{64} = \frac{280}{4096}\), which simplifies to \(\frac{35}{512}\), not equal to \(\frac{4}{64}\). Therefore, events R and S are not independent.

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

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

The possible outcomes for the sum to be 4 are (1,3), (2,2), and (3,1). Thus, \(P(S) = \frac{3}{16}\).

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

The product is odd if both numbers are odd. The possible outcomes are (1,1), (1,3), (3,1), and (3,3). Thus, \(P(T) = \frac{4}{16}\).

Now, calculate \(P(S \cap T)\):

The outcomes that satisfy both events are (1,3) and (3,1). Thus, \(P(S \cap T) = \frac{2}{16}\).

Check independence:

\(P(S) \times P(T) = \frac{3}{16} \times \frac{4}{16} = \frac{12}{256} = \frac{3}{64}\).

Since \(\frac{3}{64} \neq \frac{2}{16}\), events S and T are not independent.

For exclusivity, events are exclusive if \(P(S \cap T) = 0\).

Since \(P(S \cap T) = \frac{2}{16} \neq 0\), events S and T are not exclusive.

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 \(P(A)\):

There are 2 numbers divisible by 3 (3 and 6) and 4 numbers not divisible by 3 (1, 2, 4, 5) on a die. The probability that one die shows a number divisible by 3 and the other does not is:

\(P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}\)

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

The product is even if at least one number is even. There are 3 even numbers (2, 4, 6) on a die. The probability that the product is even is:

\(P(B) = 1 - \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{4}\)

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

\(P(A \cap B) = \frac{12}{36} = \frac{1}{3}\)

Check independence:

\(P(A) \times P(B) = \frac{4}{9} \times \frac{3}{4} = \frac{1}{3}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

For mutual exclusivity, events are mutually exclusive if \(P(A \cap B) = 0\). Since \(P(A \cap B) = \frac{1}{3} \neq 0\), events A and B are not mutually exclusive.

The probability of event A, where the white marble is in either box L or M, is given by:

\(P(A) = \frac{1}{2}\)

The probability of event B, where the red marble is to the left of both the green marble and the yellow marble, is given by:

\(P(B) = \frac{8}{24} = \frac{1}{3}\)

The probability of both events A and B occurring together is:

\(P(A \cap B) = \frac{1}{6}\)

To check for independence, we verify if:

\(P(A) \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)

Since \(P(A \cap B) = P(A) \times P(B)\), events A and B are independent.

(iii) To determine if events \(R\) and \(S\) are independent, we check if \(P(R \cap S) = P(R) \times P(S)\).

Given: \(P(R) = 0.5\), \(P(S) = 0.4\), and \(P(R \cap S) = \frac{120}{720} = \frac{1}{6}\).

Calculate \(P(R) \times P(S) = 0.5 \times 0.4 = 0.2\).

Since \(P(R \cap S) = \frac{1}{6} \neq 0.2\), events \(R\) and \(S\) are not independent.

(iv) Events \(R\) and \(S\) are exclusive if \(P(R \cap S) = 0\). However, \(P(R \cap S) = \frac{1}{6} \neq 0\), indicating there is an overlap between \(R\) and \(S\) (e.g., the combination 3, 4, 4).

Therefore, events \(R\) and \(S\) are not exclusive.

The probability of event Q, getting a total of 5, can occur with the following pairs: (1,4), (2,3), (3,2), (4,1). Thus,

\(P(Q) = \frac{4}{36} = \frac{1}{9}\).

The probability of event S, getting one low score and one high score, can occur with the following pairs: (1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,1), (5,1), (6,1), (4,2), (5,2), (6,2), (4,3), (5,3), (6,3). Thus,

\(P(S) = \frac{18}{36} = \frac{1}{2}\).

The probability of both events Q and S occurring, i.e., the intersection \(Q \cap S\), is given by the pairs: (1,4), (2,3), (3,2), (4,1). Thus,

\(P(Q \cap S) = \frac{4}{36} = \frac{1}{9}\).

To check for independence, we verify if \(P(Q \cap S) = P(Q) \times P(S)\).

\(P(Q) \times P(S) = \frac{1}{9} \times \frac{1}{2} = \frac{1}{18}\).

Since \(P(Q \cap S) = \frac{1}{9}\) and \(P(Q) \times P(S) = \frac{1}{18}\), the events are independent.

(i) The total number of countries is 170. The number of countries with a medium GDP is \(20 + 42 + 12 = 74\). Therefore, the probability is \(\frac{74}{170} = \frac{37}{85} \approx 0.435\).

(ii) The number of countries that do not have a medium GDP is \(170 - 74 = 96\). The number of countries with a low birth rate and not a medium GDP is \(3 + 35 = 38\). Therefore, the probability is \(\frac{38}{96} = \frac{19}{49} \approx 0.396\).

(iii) The probability of a country having both a high GDP and a high birth rate is 0, as there are no such countries. Therefore, the events are exclusive.

(iv) The probability of choosing a country with a medium GDP is \(\frac{74}{170} = \frac{37}{85}\). The probability of choosing a country with a medium birth rate is \(\frac{42}{96} = \frac{7}{16}\). The combined probability is \(\frac{37}{85} \times \frac{7}{16} = \frac{287}{1360} \approx 0.211\).

(i) Complete the table:

(ii) The probability of choosing the pair of low-heeled shoes with designer labels is \(\frac{1}{20}\).

(iii) The probability of choosing a pair of sports shoes is \(\frac{10}{20} = \frac{1}{2}\).

(iv) The probability of choosing a pair of high-heeled shoes given that it has a designer label is \(\frac{2}{8} = \frac{1}{4}\).

(v) To check independence, calculate:

\(P(D) = \frac{8}{20} = 0.4\)

\(P(S) = \frac{10}{20} = 0.5\)

\(P(D \cap S) = \frac{5}{20} = 0.25\)

Since \(P(D) \times P(S) = 0.4 \times 0.5 = 0.2 \neq 0.25 = P(D \cap S)\), the events are not independent.

(i) Let \(F\) be the event that a person is female, and \(W\) be the event that a person watches 'Kops are Kids'.

\(P(F) = \frac{12}{30} = 0.4\)

\(P(W) = \frac{16}{30} = 0.533\)

\(P(F \cap W') = \frac{5}{30} = 0.167\)

Using the formula for the union of two events:

\(P(F \cup W) = P(F) + P(W) - P(F \cap W')\)

\(P(F \cup W) = \frac{12}{30} + \frac{16}{30} - \frac{5}{30} = \frac{5}{6} = 0.833\)

(ii) Let \(M\) be the event that a person is male.

\(P(M) = \frac{18}{30} = 0.6\)

\(P(W) = \frac{16}{30} = 0.533\)

\(P(M \cap W) = \frac{13}{30} = 0.433\)

Check independence by comparing \(P(M \cap W)\) and \(P(M) \times P(W)\):

\(P(M) \times P(W) = \frac{18}{30} \times \frac{16}{30} = \frac{8}{25} = 0.32\)

Since \(P(M \cap W) \neq P(M) \times P(W)\), the events are not independent.

(i) Calculate the probability for each scenario where exactly two balls have the same number:

- Probability of selecting 2 from A, 2 from B, and any from C: \(\frac{1}{4} \times \frac{1}{5} \times \frac{1}{7} = \frac{1}{140}\)

- Probability of selecting 8 from A, 8 from B, and any from C: \(\frac{1}{4} \times \frac{2}{5} \times \frac{3}{7} = \frac{3}{70}\)

- Probability of selecting 8 from A, any from B, and 8 from C: \(\frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}\)

- Probability of selecting any from A, 8 from B, and 8 from C: \(\frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}\)

Summing these probabilities gives \(\frac{47}{140}\).

(ii) Given that exactly two of the selected balls have the same number, the probability that they are both numbered 2 is:

\(\frac{\frac{1}{28}}{\frac{47}{140}} = \frac{5}{47}\)

(iii) Calculate \(P(X) = \frac{47}{140}\) and \(P(Y) = \frac{1}{4}\). Check independence:

\(P(X \cap Y) = \frac{1}{28}\)

Since \(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.

(i) (a) To get a final score of 12, the second throw must be 6 and the first throw must also be 6, since \(6 + 6 = 12\). The probability of this is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).

(b) To get a final score of 5, the possible outcomes are (1,4), (2,3), (3,2), (4,1) where the second throw is not 5. The probability is \(\frac{4}{36} = \frac{1}{9}\).

(ii) Calculate probabilities:

\(P(A) = \frac{1}{6}\)

\(P(B) = \frac{5}{36}\) for pairs (1,5), (2,4), (3,3), (4,2), (5,1)

\(P(C) = 1 - P(O, O) = \frac{3}{4}\)

Check independence:

\(P(A \cap B) = \frac{1}{36}\) which is not equal to \(P(A) \times P(B)\)

\(P(A \cap C) = \frac{3}{36}\) which is not equal to \(P(A) \times P(C)\)

\(P(B \cap C) = \frac{2}{36}\) which is not equal to \(P(B) \times P(C)\)

None of the events are independent.

(i) To find \(P(Q)\), we need the pairs \((a, b)\) such that \(a \times b = 24\). The possible pairs are \((2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2)\). There are 6 favorable outcomes. The total number of outcomes when two dice are thrown is \(12 \times 12 = 144\). Thus, \(P(Q) = \frac{6}{144} = \frac{1}{24}\).

(ii) For \(P(R)\), both numbers must be greater than 8. The numbers greater than 8 are 9, 10, 11, and 12. Thus, there are 4 choices for each die, giving \(4 \times 4 = 16\) favorable outcomes. Therefore, \(P(R) = \frac{16}{144} = \frac{1}{9}\).

(iii) Events \(Q\) and \(R\) are exclusive if they cannot occur simultaneously. Since \(R\) requires both numbers to be greater than 8, and none of the pairs for \(Q\) satisfy this, \(P(R \cap Q) = 0\). Thus, \(R\) and \(Q\) are exclusive.

(iv) Events \(Q\) and \(R\) are independent if \(P(R \cap Q) = P(R) \times P(Q)\). Since \(P(R \cap Q) = 0\) and \(P(R) \times P(Q) \neq 0\), they are not independent.

(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.

(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.