Write down the value of ${\text{P}}(X = 1)$ and show that ${\text{P}}(X = 2) = \frac{1}{6}$.

Show that the probability generating function for X is given by

$$G(t) = \frac{{2t + {t^2}}}{{6 - 3{t^2}}}.$$

Hence determine ${\text{E}}(X)$.

${\text{P}}(X = 1) = \frac{1}{3}$     A1

${\text{P}}(X = 2) = \frac{2}{3} \times \frac{1}{4}$     A1

$= \frac{1}{6}$     AG

[2 marks]

$G(t) = \frac{1}{3}t + \frac{2}{3} \times \frac{1}{4}{t^2} + \frac{2}{3} \times \frac{3}{4} \times \frac{1}{3}{t^3} + \frac{2}{3} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{4}{t^4} +  \ldots$     M1A1

$= \frac{1}{3}t\left( {1 + \frac{1}{2}{t^2} +  \ldots } \right) + \frac{1}{6}{t^2}\left( {1 + \frac{1}{2}{t^2} +  \ldots } \right)$     M1A1

$= \frac{{\frac{t}{3}}}{{1 - \frac{{{t^2}}}{2}}} + \frac{{\frac{{{t^2}}}{6}}}{{1 - \frac{{{t^2}}}{2}}}$     A1A1

$= \frac{{2t + {t^2}}}{{6 - 3{t^2}}}$     AG

[6 marks]

$G'(t) = \frac{{(2 + 2t)(6 - 3{t^2}) + 6t(2t + {t^2})}}{{{{(6 - 3{t^2})}^2}}}$     M1A1

${\text{E}}(X) = G'(1) = \frac{{10}}{3}$     M1A1

[4 marks]