Question
A species of bird can nest in two seasons: Spring and Summer.
The probability of nesting in Spring is $k$.
The probability of nesting in Summer is $\frac{k}{2}$.
This is shown in the following tree diagram.
(a) Complete the tree diagram to show the probabilities of not nesting in each season. Write your answers in terms of $k$.
It is known that the probability of not nesting in Spring and not nesting in Summer is $\frac{5}{9}$.
(b) (i) Show that $9k^2-27k+8=0$.
(ii) Both $k=\frac{1}{3}$ and $k=\frac{8}{3}$ satisfy $9k^2-27k+8=0$. State why $k=\frac{1}{3}$ is the only valid solution.
▶️Answer/Explanation
Detailed solution
(a)\[ \text{probability(nesting)} + \text{probability(not nesting)} = 1 \] For spring season, \[ k + \text{probability(not nesting)} = 1 \] \[ \text{probability(not nesting)} = 1 – k \]
For summer, \[ \left(\frac{k}{2}\right) + \text{probability(not nesting)} = 1 \] \[ \text{probability(not nesting)} = \left(1 – \frac{k}{2} \right) \]
(b) (i) Since the two given events are independent} Given, \[ (\text{probability of not nesting in Spring}) \times (\text{probability of not nesting in Summer}) = \frac{5}{9} \] \[ (1 – k) \left( 1 – \frac{k}{2} \right) = \frac{5}{9} \] Expanding the LHS, \[ 1 – k – \frac{k}{2} + \frac{k^2}{2} = \frac{5}{9} \] Multiplying throughout by 18 to eliminate fractions, \[ 18 – 18k – 9k + 9k^2 = 10 \] \[ 9k^2 – 27k + 8 = 0 \]
Hence proved.
(ii) Since the probability of an event can never be greater than 1, therefore, \[ k = \frac{1}{3} \] is the only valid solution, as \[ \frac{8}{3} > 1 \]
Mark Scheme-
$1-k$ for Spring
$1-\frac{k}{2}$ for both summers
(b) (i) multiplying the two correct branches
$(1-k)(1-\frac{k}{2})$
attempt to expand and equate to $\frac{5}{9}$
$1-k-\frac{k}{2}+\frac{k^{2}}{2}=\frac{5}{9}$
$18-18k-9k+9k^{2}=10$ OR $\frac{k^{2}}{2}-\frac{3k}{2}+\frac{4}{9}=0$ OR $\frac{9k^{2}}{2}-\frac{27k}{2}+4=0$
$9k^{2}-27k+8=0$
(ii) $(k=\frac{1}{3}$ is the only valid solution as) $\frac{8}{3}>1$