CAIE A-Level Further Math AS 4.5 Probability Generating Functions Questions

$\mathrm{P}(X=r)=p(1-p)^{r-1}=p q^{r-1}$
Implied by $\mathrm{G}(t)=\sum p q^{r-1} t^{r}$.

$\mathrm{G}(t)=\sum p t(q t)^{r-1}$
(Infinite summation) with common ratio ( q t ) and first
term or common factor pt indicated or GP identified.

$\mathrm{G}(t)=p t\left(\frac{1}{1-q t}\right)=\frac{p t}{1-q t}$
AG
Convincingly obtained.

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$\mathrm{G}^{\prime}(t)=p q t(1-q t)^{-2}+p(1-q t)^{-1}\left[=p(1-q t)^{-2}\right]$
Attempt to differentiate as a product/quotient.

$\mathrm{G}^{\prime \prime}(t)=2 p q^{2} t(1-q t)^{-3}+2 p q(1-q t)^{-2} \quad\left[=2 q p(1-q t)^{-3}\right]$
Attempt to differentiate their $\mathrm{G}^{\prime}(t)$.

$\mathrm{G}^{\prime}(1)=p(1-q)^{-2}=\frac{1}{p}, \quad \mathrm{G}^{\prime \prime}(1)=\frac{2 q}{P^{2}}$
Substitute t=1 into their $\mathrm{G}^{\prime}(t)$ and $\mathrm{G}^{\prime \prime}(t)$ and use formula
for $\operatorname{Var}(X)$ (at some point).

$\operatorname{Var}(X)=\frac{2 q}{P^{2}}+\frac{1}{p}-\left(\frac{1}{p}\right)^{2}$
(1-q) replaced by p in denominator throughout, and
some cancellation seen (at some point).

$\frac{q}{p^{2}}$
AG
Convincingly obtained.

5

$\mathrm{G}_{X}(t)=\frac{\frac{1}{6} t}{1-\frac{5}{6} t} \quad\left[=\frac{t}{6-5 t}\right]$
Implied by correct final answer.

$\mathrm{G}_{z}(t)=\left(\mathrm{G}_{X}(t)\right)^{2}=\left(\frac{t}{6-5 t}\right)^{2}$
Allow $\left(\frac{\frac{1}{6} t}{1-\frac{5}{6} t}\right)^{2}$.

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