(i) The probability that the person chosen is from country A is given by the ratio of the number of people in country A to the total number of people:

\(P(A) = \frac{3}{15} = 0.2\)

(ii) To find the probability that the person chosen does not have sugar in their tea, we calculate the probabilities for both countries and sum them:

In country A, 70% do not have sugar: \(0.7 \times 3 \text{ million} = 2.1 \text{ million}\)

In country B, 35% do not have sugar: \(0.35 \times 12 \text{ million} = 4.2 \text{ million}\)

Total without sugar: \(2.1 + 4.2 = 6.3 \text{ million}\)

Probability: \(P(\text{not S}) = \frac{6.3}{15} = 0.42\)

(iii) Given that the person does not have sugar, the probability they are from country B is:

\(P(B | \text{not S}) = \frac{0.8 \times 0.35}{0.42} = 0.667\)

(i) The probability of taking a white paper clip from box A is \(\frac{1}{6}\). After transferring this white paper clip to box B, box B will have 7 red and 3 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{7}{10}\). Therefore, the probability of both events occurring is:

\(P(W, R) = \frac{1}{6} \times \frac{7}{10} = \frac{7}{60} \approx 0.117\)

(ii) The probability of taking a red paper clip from box A is \(\frac{5}{6}\). After transferring this red paper clip to box B, box B will have 8 red and 2 white paper clips. The probability of then taking a red paper clip from box B is \(\frac{8}{10}\). Therefore, the probability of both events occurring is:

\(P(R, R) = \frac{5}{6} \times \frac{8}{10} = \frac{40}{60}\)

The total probability of taking a red paper clip from box B is the sum of the probabilities of the two mutually exclusive events (taking a red from A and then a red from B, or taking a white from A and then a red from B):

\(P(\text{red}) = \frac{40}{60} + \frac{7}{60} = \frac{47}{60} \approx 0.783\)

(iii) The probability that the paper clip taken from box A was red, given that the paper clip from box B is red, is calculated using conditional probability:

\(P(R | R) = \frac{P(R \cap R)}{P(R)} = \frac{\frac{40}{60}}{\frac{47}{60}} = \frac{40}{47} \approx 0.851\)

Let \(A\) be the event that Jamie attends the training session, and \(T\) be the event that Jamie is chosen for the team.

Given: \(P(A) = 0.5\), \(P(T|A) = 1\), \(P(T|A^c) = 0.6\).

(i) The probability that Jamie is chosen for the team is:

\(P(T) = P(T|A)P(A) + P(T|A^c)P(A^c) = 1 \times 0.5 + 0.6 \times 0.5 = 0.5 + 0.3 = 0.8\)

(ii) The conditional probability that Jamie attended the training session, given that he was chosen for the team, is:

\(P(A|T) = \frac{P(T|A)P(A)}{P(T)} = \frac{1 \times 0.5}{0.8} = \frac{0.5}{0.8} = 0.625\)

Thus, \(P(A|T) = 0.625\) or \(\frac{5}{8}\).

(i) To find the probability that a taxi arrives late, we use the law of total probability. Let L be the event that a taxi arrives late. We have:

\(P(L) = P(A) \cdot P(L|A) + P(B) \cdot P(L|B) + P(C) \cdot P(L|C)\)

Substituting the given probabilities:

\(P(L) = 0.5 \times 0.04 + 0.3 \times 0.06 + 0.2 \times 0.17\)

\(P(L) = 0.02 + 0.018 + 0.034 = 0.072\)

Thus, the probability that a taxi arrives late is 0.072.

(ii) To find the conditional probability that Andrea rang company B given that the taxi arrives late, we use Bayes' theorem:

\(P(B|L) = \frac{P(B) \cdot P(L|B)}{P(L)}\)

Substituting the known values:

\(P(B|L) = \frac{0.3 \times 0.06}{0.072}\)

\(P(B|L) = \frac{0.018}{0.072} = 0.25\)

Thus, the conditional probability that she rang company B is 0.25.

(i) The tree diagram is constructed as follows:

• First serve in: Probability = 0.65
• Win: Probability = 0.9
• Lose: Probability = 0.1
• First serve out: Probability = 0.35
• Second serve in: Probability = 0.8
• Win: Probability = 0.6
• Lose: Probability = 0.4
• Second serve out: Probability = 0.2
• Lose: Probability = 1

(ii) The probability that Don loses the point is calculated by considering all paths that lead to losing:

\(P(\text{Lose}) = 0.65 \times 0.1 + 0.35 \times 0.8 \times 0.4 + 0.35 \times 0.2 \times 1\)

\(= 0.065 + 0.112 + 0.07 = 0.247\)

(iii) The conditional probability that Don’s first serve went into the correct area, given that he loses the point, is:

\(P(\text{First in | Lose}) = \frac{0.65 \times 0.1}{0.247}\)

\(= \frac{0.065}{0.247} \approx 0.263\) or \(\frac{5}{19}\)