| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Derive standard distribution PGF |
| Difficulty | Standard +0.3 This is a standard Further Maths question on PGFs requiring derivation of the geometric distribution PGF from first principles, using differentiation to find variance, and applying the independence property. All techniques are routine for Further Statistics students with no novel problem-solving required, though it involves multiple steps and careful algebraic manipulation. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=r) = p(1-p)^{r-1} = pq^{r-1}\) | B1 | Implied by \(G(t) = \sum pq^{r-1}t^r\) |
| \(G(t) = \sum pt(qt)^{r-1}\) | M1 | Infinite summation with common ratio \((qt)\) and first term or common factor \(pt\) indicated or GP identified |
| \(G(t) = pt\left(\dfrac{1}{1-qt}\right) = \dfrac{pt}{1-qt}\) | A1 | AG, convincingly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G'(t) = pqt(1-qt)^{-2} + p(1-qt)^{-1}\ \left[= p(1-qt)^{-2}\right]\) | M1 | Attempt to differentiate as a product/quotient |
| \(G''(t) = 2pq^2t(1-qt)^{-3} + 2pq(1-qt)^{-2}\ \left[= 2qp(1-qt)^{-3}\right]\) | M1 | Attempt to differentiate their \(G'(t)\) |
| \(G'(1) = p(1-q)^{-2} = \frac{1}{p}\); \(G''(1) = \frac{2q}{p^2}\) | M1 | Substitute \(t=1\) into their \(G'(t)\) and \(G''(t)\) and use formula for \(\text{Var}(X)\) |
| \(\text{Var}(X) = \frac{2q}{p^2} + \frac{1}{p} - \left(\frac{1}{p}\right)^2\) | M1 | \((1-q)\) replaced by \(p\) in denominator throughout, and some cancellation seen |
| \(\dfrac{q}{p^2}\) | A1 | AG, convincingly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G_X(t) = \dfrac{\frac{1}{6}t}{1-\frac{5}{6}t} \left[= \dfrac{t}{6-5t}\right]\) | B1 | Implied by correct final answer. |
| \(G_Z(t) = (G_X(t))^2 = \left(\dfrac{t}{6-5t}\right)^2\) | B1 | Allow \(\left(\dfrac{\frac{1}{6}t}{1-\frac{5}{6}t}\right)^2\) |
| 2 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=r) = p(1-p)^{r-1} = pq^{r-1}$ | B1 | Implied by $G(t) = \sum pq^{r-1}t^r$ |
| $G(t) = \sum pt(qt)^{r-1}$ | M1 | Infinite summation with common ratio $(qt)$ and first term or common factor $pt$ indicated or GP identified |
| $G(t) = pt\left(\dfrac{1}{1-qt}\right) = \dfrac{pt}{1-qt}$ | A1 | AG, convincingly obtained |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G'(t) = pqt(1-qt)^{-2} + p(1-qt)^{-1}\ \left[= p(1-qt)^{-2}\right]$ | M1 | Attempt to differentiate as a product/quotient |
| $G''(t) = 2pq^2t(1-qt)^{-3} + 2pq(1-qt)^{-2}\ \left[= 2qp(1-qt)^{-3}\right]$ | M1 | Attempt to differentiate their $G'(t)$ |
| $G'(1) = p(1-q)^{-2} = \frac{1}{p}$; $G''(1) = \frac{2q}{p^2}$ | M1 | Substitute $t=1$ into their $G'(t)$ and $G''(t)$ and use formula for $\text{Var}(X)$ |
| $\text{Var}(X) = \frac{2q}{p^2} + \frac{1}{p} - \left(\frac{1}{p}\right)^2$ | M1 | $(1-q)$ replaced by $p$ in denominator throughout, and some cancellation seen |
| $\dfrac{q}{p^2}$ | A1 | AG, convincingly obtained |
# Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G_X(t) = \dfrac{\frac{1}{6}t}{1-\frac{5}{6}t} \left[= \dfrac{t}{6-5t}\right]$ | B1 | Implied by correct final answer. |
| $G_Z(t) = (G_X(t))^2 = \left(\dfrac{t}{6-5t}\right)^2$ | B1 | Allow $\left(\dfrac{\frac{1}{6}t}{1-\frac{5}{6}t}\right)^2$ |
| | **2** | |
---5 The random variable $X$ has the geometric distribution $\operatorname { Geo } ( p )$.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability generating function of $X$ is $\frac { \mathrm { pt } } { 1 - \mathrm { qt } }$, where $\mathrm { q } = 1 - \mathrm { p }$.
\item Use the probability generating function of $X$ to show that $\operatorname { Var } ( X ) = \frac { \mathrm { q } } { \mathrm { p } ^ { 2 } }$.\\
Kenny throws an ordinary fair 6-sided dice repeatedly. The random variable $X$ is the number of throws that Kenny takes in order to obtain a 6 . The random variable $Z$ denotes the sum of two independent values of $X$.
\item Find the probability generating function of $Z$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q5 [10]}}