(i) To find the probability of obtaining a total score of 5, we list all possible combinations of scores on the three dice that sum to 5:

• (1, 1, 3)
• (1, 2, 2)
• (1, 3, 1)
• (2, 1, 2)
• (2, 2, 1)
• (3, 1, 1)

There are 6 combinations. Since each die has 6 faces, the total number of possible outcomes when throwing three dice is \(6^3 = 216\). Thus, the probability is \(\frac{6}{216} = \frac{1}{36}\).

(ii) To find the probability of obtaining a total score of 7, we list all possible combinations of scores on the three dice that sum to 7:

• (1, 2, 4) and permutations: 6 combinations
• (1, 3, 3) and permutations: 3 combinations
• (2, 2, 3) and permutations: 3 combinations
• (1, 1, 5) and permutations: 3 combinations

There are 15 combinations. Thus, the probability is \(\frac{15}{216} = \frac{5}{72}\).

(a) Let the probability that a student sings in the choir be denoted by P(C). We know that:

\(P(not C) = 0.58\)

\(Therefore, P(C) = 1 - 0.58 = 0.42\)

\(The probability that a student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that a student does not play in the band and sings in the choir is 0.4 × x.

Thus, the total probability of singing in the choir is:

\(0.18 + 0.4x = 0.42\)

Solving for x:

\(0.4x = 0.42 - 0.18\)

\(0.4x = 0.24\)

\(x = \frac{0.24}{0.4} = 0.6\)

\((b) The probability that one student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that both students play in the band and both sing in the choir is:

\((0.18)^2 = 0.0324\)

The number of students at Canton college who prefer hockey is 56, as shown in the table.

The total number of students surveyed is 500.

The probability that a randomly chosen student is at Canton college and prefers hockey is given by the ratio of the number of Canton hockey students to the total number of students:

\(\text{Probability} = \frac{56}{500}\)

This fraction can be simplified to \(\frac{14}{125}\).

Alternatively, the decimal form of the probability is 0.112.

Let the probability that Shona is late be the complement of the probability that she is not late. Therefore, the probability that she is late is:

\(1 - 0.63 = 0.37\)

The probability that Shona is late if she catches the bus is:

\(0.8 imes 0.4 = 0.32\)

The probability that Shona is late if she cycles is:

0.2 imes x

Thus, the total probability that Shona is late is:

\(0.32 + 0.2x = 0.37\)

Solving for x:

\(0.2x = 0.37 - 0.32\)

\(0.2x = 0.05\)

\(x = rac{0.05}{0.2} = 0.25\)

The probability of choosing a plum from Jameel's box is \(\frac{5}{8}\) because there are 5 plums out of a total of 8 fruits.

The probability of choosing a plum from Rosa's box is \(\frac{x}{x+6}\) because there are x plums out of a total of \(x + 6\) fruits.

The probability that both fruits chosen are plums is given by:

\(\frac{5}{8} \times \frac{x}{x+6} = \frac{1}{4}\)

To solve for x, multiply both sides by \(8(x+6)\):

\(5x = 2(x+6)\)

Expand and simplify:

\(5x = 2x + 12\)

Subtract \(2x\) from both sides:

\(3x = 12\)

Divide both sides by 3:

\(x = 4\)