Complete the tree diagram for probabilities of not nesting.

For Spring: Probability of nesting = \( k \), so probability of not nesting = \( 1 – k \).

For Summer: Probability of nesting = \( \frac{k}{2} \), so probability of not nesting = \( 1 – \frac{k}{2} \).

Answer: Spring: \( 1 – k \), Summer: \( 1 – \frac{k}{2} \)

(i) Show that \( 9k^2 – 27k + 8 = 0 \).

Probability of not nesting in Spring and not nesting in Summer = \( \frac{5}{9} \).

Since events are independent: \( (1 – k) \cdot \left(1 – \frac{k}{2}\right) = \frac{5}{9} \).

Expand: \[ 1 – k – \frac{k}{2} + \frac{k^2}{2} = \frac{5}{9} \]

\[ 1 – \frac{3k}{2} + \frac{k^2}{2} = \frac{5}{9} \]

Multiply by 18: \[ 18 – 27k + 9k^2 = 10 \]

\[ 9k^2 – 27k + 18 – 10 = 0 \]

\[ 9k^2 – 27k + 8 = 0 \]

(ii) State why \( k = \frac{1}{3} \) is the only valid solution.

Solve: \( 9k^2 – 27k + 8 = 0 \).

Factorize: \[ (3k – 1)(3k – 8) = 0 \]

\[ k = \frac{1}{3}, \frac{8}{3} \]

Since \( k \) is a probability, \( 0 \leq k \leq 1 \). Also, Summer probability \( \frac{k}{2} \leq 1 \), so \( k \leq 2 \).

\[ k = \frac{8}{3} > 1 \], which is invalid.

Answer: (i) \( 9k^2 – 27k + 8 = 0 \) (shown), (ii) \( k = \frac{1}{3} \) is valid as \( \frac{8}{3} > 1 \).