Let \( R \) be “it rains” and \( W \) be “the ‘Tigers’ soccer team win”.

(a) [1 mark]

EITHER
Tree diagram with first branches: \( P(R) = \frac{2}{5} \), \( P(R’) = \frac{3}{5} \); second branches: \( P(W | R) = \frac{2}{7} \), \( P(W’ | R) = \frac{5}{7} \), \( P(W | R’) = \frac{4}{7} \), \( P(W’ | R’) = \frac{3}{7} \) (A1).
OR
Correct structure with probabilities labeled on branches as shown in the image: (A1).

(b) [2 marks]

EITHER
\( P(W) = P(R \cap W) + P(R’ \cap W) = \left( \frac{2}{5} \times \frac{2}{7} \right) + \left( \frac{3}{5} \times \frac{4}{7} \right) = \frac{4}{35} + \frac{12}{35} = \frac{16}{35} \) (M1A1).
OR
Use tree diagram to sum winning paths: \( \frac{2}{5} \times \frac{2}{7} + \frac{3}{5} \times \frac{4}{7} = \frac{16}{35} \) (M1A1).

(c) [2 marks]

EITHER
\( P(R | W) = \frac{P(R \cap W)}{P(W)} = \frac{\frac{2}{5} \times \frac{2}{7}}{\frac{16}{35}} = \frac{\frac{4}{35}}{\frac{16}{35}} = \frac{4}{16} = \frac{1}{4} \) (M1A1).
OR
Use conditional probability formula with tree diagram values: \( \frac{P(R \cap W)}{P(R \cap W) + P(R’ \cap W)} = \frac{1}{4} \) (M1A1).