(i) The tree diagram is drawn with the following branches:

• First set of branches for the type of message: text (0.3), email (0.2), social media (0.5).
• Second set of branches for replies:
• Text: reply (0.4), no reply (0.6).
• Email: reply (0.15), no reply (0.85).
• Social media: reply (0.6), no reply (0.4).

(ii) To find the probability that the message was an email given no immediate reply, use Bayes' theorem:

\(P(\text{email} \mid \text{NR}) = \frac{P(\text{email} \cap \text{NR})}{P(\text{NR})}\)

Calculate \(P(\text{email} \cap \text{NR}) = 0.2 \times 0.85 = 0.17\).

Calculate \(P(\text{NR}) = 0.3 \times 0.6 + 0.2 \times 0.85 + 0.5 \times 0.4 = 0.18 + 0.17 + 0.2 = 0.55\).

Thus, \(P(\text{email} \mid \text{NR}) = \frac{0.17}{0.55} = \frac{17}{55} \approx 0.309\)

(i) The probability that Tamar wears red socks is calculated by considering both cases: wearing a blue suit and wearing a grey suit.

Probability of wearing red socks with a blue suit: \(0.6 \times 0.2 = 0.12\).

Probability of wearing red socks with a grey suit: \(0.4 \times 0.32 = 0.128\).

Total probability of wearing red socks: \(0.12 + 0.128 = 0.248\).

Expressed as a fraction: \(\frac{31}{125}\).

(ii) We need to find the probability that Tamar is wearing a grey suit given that he is not wearing red socks.

Probability of not wearing red socks with a blue suit: \(0.6 \times 0.8 = 0.48\).

Probability of not wearing red socks with a grey suit: \(0.4 \times 0.68 = 0.272\).

Total probability of not wearing red socks: \(0.48 + 0.272 = 0.752\).

Using Bayes' theorem, the probability of wearing a grey suit given not wearing red socks is:

\(\frac{0.272}{0.752} = \frac{17}{47}\).

(i) The tree diagram is completed as follows:

• First ball: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{5}{8}\).
• Second ball after red: Probability of red (R) = \(\frac{2}{8}\), Probability of blue (B) = \(\frac{5}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).
• Second ball after blue: Probability of red (R) = \(\frac{3}{8}\), Probability of blue (B) = \(\frac{4}{8}\), Probability of yellow (Y) = \(\frac{1}{8}\).

(ii) Probability that the two balls taken are the same colour:

\(P(RR) = \frac{3}{8} \times \frac{2}{8} = \frac{3}{32}\)

\(P(BB) = \frac{5}{8} \times \frac{4}{8} = \frac{5}{16}\)

Total probability = \(\frac{3}{32} + \frac{5}{16} = \frac{13}{32}\) or 0.406

(iii) Probability that the first ball taken is red, given that the second ball taken is blue:

\(P(RB) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}\)

\(P(B) = \frac{3}{8} \times \frac{5}{8} + \frac{5}{8} \times \frac{4}{8} = \frac{35}{64}\)

\(P(R|B) = \frac{P(RB)}{P(B)} = \frac{15/64}{35/64} = \frac{3}{7}\) or 0.429

First, calculate the total number of girls:

\(15 + 12 + 37 = 64\).

Next, calculate the number of girls not studying Drama:

\(15 + 12 = 27\).

The probability that a randomly chosen student is not studying Drama, given that the student is a girl, is:

\(\frac{27}{64}\).

(i) Let the probability of being a Beginner be 0.5. The probability of being a Beginner is given by:

\(0.7x + 0.45(1-x) = 0.5\)

Solving for \(x\):

\(0.7x + 0.45 - 0.45x = 0.5\)

\(0.25x = 0.05\)

\(x = 0.2\)

(ii) The probability of being an Advanced swimmer is also 0.5. The probability of being an Advanced swimmer and male is:

\(0.3x = 0.3 \times 0.2 = 0.06\)

The probability of being an Advanced swimmer is:

\(0.5 = 0.3 \times 0.2 + 0.55 \times 0.8\)

The probability that a randomly chosen Advanced swimmer is male is:

\(\frac{0.06}{0.5} = 0.12 = \frac{3}{25}\)