| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Multiple independent coins/dice |
| Difficulty | Standard +0.8 This is a standard Further Maths probability generating functions question requiring knowledge of PGF manipulation, independence properties, and extracting probabilities from polynomials. While it involves multiple steps (finding individual PGFs, combining them, calculating variance, finding mode), each step follows routine procedures taught in UFM Statistics. The biased dice adds minor complexity but no novel insight is required—it's a textbook-style multi-part question slightly above average difficulty due to the topic being Further Maths content. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_X(t) = \frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2\) | M1 A1 | 2 probabilities correct, in a polynomial |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Z(t) = \left(\frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2\right)\left(\frac{25}{36} + \frac{10}{36}t + \frac{1}{36}t^2\right)\) | M1 | Second PGF correct and multiplied by part (a) |
| \(\frac{1}{576}\left(225 + 240t + 94t^2 + 16t^3 + t^4\right)\) | M1 | Obtains a quartic polynomial |
| Or \(\frac{25}{64} + \frac{5}{12}t + \frac{47}{288}t^2 + \frac{1}{36}t^3 + \frac{1}{576}t^4\) | A1 | Note: \(\frac{1}{576}(t+3)^2(t+5)^2\) scores M1M1A0 |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Z'(t) = \frac{1}{576}\left(240 + 188t + 48t^2 + 4t^3\right)\) | M1 | Differentiate twice |
| \(G_Z''(t) = \frac{1}{576}\left(188 + 96t + 12t^2\right)\) | ||
| \(\text{Var}(Z) = \frac{1}{576}(188 + 96 + 12) + \frac{5}{6} - \frac{25}{36}\) | M1 | Use correct formula |
| \(\frac{47}{72}\) | A1 | |
| OR: \(\text{Var}(Z) = 2 \times \frac{1}{4} \times \frac{3}{4} + 2 \times \frac{1}{6} \times \frac{5}{6}\) | M1 M1 | One term correct; Two terms present and added |
| \(\frac{47}{72}\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 | B1 FT | FT their power with greatest coefficient in part (b) |
| 1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_X(t) = \frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2$ | M1 A1 | 2 probabilities correct, in a polynomial |
| | **2** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z(t) = \left(\frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2\right)\left(\frac{25}{36} + \frac{10}{36}t + \frac{1}{36}t^2\right)$ | M1 | Second PGF correct and multiplied by part (a) |
| $\frac{1}{576}\left(225 + 240t + 94t^2 + 16t^3 + t^4\right)$ | M1 | Obtains a quartic polynomial |
| Or $\frac{25}{64} + \frac{5}{12}t + \frac{47}{288}t^2 + \frac{1}{36}t^3 + \frac{1}{576}t^4$ | A1 | Note: $\frac{1}{576}(t+3)^2(t+5)^2$ scores M1M1A0 |
| | **3** | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z'(t) = \frac{1}{576}\left(240 + 188t + 48t^2 + 4t^3\right)$ | M1 | Differentiate twice |
| $G_Z''(t) = \frac{1}{576}\left(188 + 96t + 12t^2\right)$ | | |
| $\text{Var}(Z) = \frac{1}{576}(188 + 96 + 12) + \frac{5}{6} - \frac{25}{36}$ | M1 | Use correct formula |
| $\frac{47}{72}$ | A1 | |
| OR: $\text{Var}(Z) = 2 \times \frac{1}{4} \times \frac{3}{4} + 2 \times \frac{1}{6} \times \frac{5}{6}$ | M1 M1 | One term correct; Two terms present and added |
| $\frac{47}{72}$ | A1 | |
| | **3** | |
---
## Question 5(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 | B1 FT | FT their power with greatest coefficient in part (b) |
| | **1** | |5 A 6 -sided dice, $A$, with faces numbered $1,2,3,4,5,6$ is biased so that the probability of throwing a 6 is $\frac { 1 } { 4 }$. The random variable $X$ is the number of 6s obtained when dice $A$ is thrown twice.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function of $X$.\\
A second dice, $B$, with faces numbered $1,2,3,4,5,6$ is unbiased. The random variable $Y$ is the number of 6s obtained when dice $B$ is thrown twice.
The random variable $Z$ is the total number of 6s obtained when both dice are thrown twice.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.
\item Find $\operatorname { Var } ( Z )$.
\item Use the probability generating function of $Z$ to find the most probable value of $Z$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q5 [9]}}