| 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 | Selection without replacement scenarios |
| Difficulty | Challenging +1.2 This is a structured Further Maths statistics question on PGFs with clear scaffolding. Part (a) requires computing hypergeometric probabilities and forming the PGF—standard technique. Part (b) uses the given PGF for Y and multiplies with the PGF from (a), requiring polynomial expansion. Part (c) applies the standard variance formula G''(1) + G'(1) - [G'(1)]². While this is Further Maths content (inherently harder), the question follows a predictable template with no novel insights required, making it slightly above average difficulty overall. |
| 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 |
| \(P(3R) = \frac{120}{504}\), \(P(2R) = \frac{270}{504}\), \(P(1R) = \frac{108}{504}\), \(P(0R) = \frac{6}{504}\) | B1 | 2 correct probabilities. |
| \(G_X(t) = \frac{1}{504}\left(6 + 108t + 270t^2 + 120t^3\right) = \frac{1}{84}\left(1 + 18t + 45t^2 + 20t^3\right)\) | M1 | Cubic polynomial with their probabilities. |
| \(\frac{1}{84} + \frac{3}{14}t + \frac{15}{28}t^2 + \frac{5}{21}t^3\) | A1 | Correct. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Z(t) = \frac{1}{84}\left(1 + 18t + 45t^2 + 20t^3\right) \times \frac{1}{12}\left(1 + 6t + 5t^2\right)\) | M1 | Attempt to multiply out the brackets. |
| \(\frac{1}{1008}\left(1 + 24t + 158t^2 + 380t^3 + 345t^4 + 100t^5\right)\) | M1 | Obtain quintic polynomial (may not be simplified). |
| \(\frac{1}{1008} + \frac{1}{42}t + \frac{79}{504}t^2 + \frac{95}{252}t^3 + \frac{115}{336}t^4 + \frac{25}{252}t^5\) | A1 | Correct. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G' = \frac{1}{1008}\left(24 + 316t + 1140t^2 + 1380t^3 + 500t^4\right)\); \(G'' = \frac{1}{1008}\left(316 + 2280t + 4140t^2 + 2000t^3\right)\) | M1 | Differentiate twice. May not see derivatives in terms of \(t\). |
| \(\left[E(Z) = \frac{3360}{1008}\right] = \frac{10}{3}\) | B1 | Seen or implied, NFWW (e.g. sampling with replacement). |
| \(\text{Var}(Z) = \frac{8736}{1008} + \frac{10}{3} - \left(\frac{10}{3}\right)^2 \left[= \frac{26}{3} - \frac{10}{3} - \frac{100}{9}\right]\) | M1 | Use formula with their \(G''(1)\) and \(G'(1)\). |
| \(\frac{8}{9}\) | A1 | Or 0.889 |
| 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(3R) = \frac{120}{504}$, $P(2R) = \frac{270}{504}$, $P(1R) = \frac{108}{504}$, $P(0R) = \frac{6}{504}$ | B1 | 2 correct probabilities. |
| $G_X(t) = \frac{1}{504}\left(6 + 108t + 270t^2 + 120t^3\right) = \frac{1}{84}\left(1 + 18t + 45t^2 + 20t^3\right)$ | M1 | Cubic polynomial with their probabilities. |
| $\frac{1}{84} + \frac{3}{14}t + \frac{15}{28}t^2 + \frac{5}{21}t^3$ | A1 | Correct. |
| | **3** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z(t) = \frac{1}{84}\left(1 + 18t + 45t^2 + 20t^3\right) \times \frac{1}{12}\left(1 + 6t + 5t^2\right)$ | M1 | Attempt to multiply out the brackets. |
| $\frac{1}{1008}\left(1 + 24t + 158t^2 + 380t^3 + 345t^4 + 100t^5\right)$ | M1 | Obtain quintic polynomial (may not be simplified). |
| $\frac{1}{1008} + \frac{1}{42}t + \frac{79}{504}t^2 + \frac{95}{252}t^3 + \frac{115}{336}t^4 + \frac{25}{252}t^5$ | A1 | Correct. |
| | **3** | |
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G' = \frac{1}{1008}\left(24 + 316t + 1140t^2 + 1380t^3 + 500t^4\right)$; $G'' = \frac{1}{1008}\left(316 + 2280t + 4140t^2 + 2000t^3\right)$ | M1 | Differentiate twice. May not see derivatives in terms of $t$. |
| $\left[E(Z) = \frac{3360}{1008}\right] = \frac{10}{3}$ | B1 | Seen or implied, NFWW (e.g. sampling with replacement). |
| $\text{Var}(Z) = \frac{8736}{1008} + \frac{10}{3} - \left(\frac{10}{3}\right)^2 \left[= \frac{26}{3} - \frac{10}{3} - \frac{100}{9}\right]$ | M1 | Use formula with their $G''(1)$ and $G'(1)$. |
| $\frac{8}{9}$ | A1 | Or 0.889 |
| | **4** | |3 Toby has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 3 marbles from the bag, without replacement. The random variable $X$ is the number of red marbles that Toby obtains.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function of $X$.\\
Ling also has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 2 marbles from his bag, without replacement. The random variable $Y$ is the number of red marbles that Ling obtains. It is given that the probability generating function of $Y$ is $\frac { 1 } { 12 } \left( 1 + 6 t + 5 t ^ { 2 } \right)$.
The random variable $Z$ is the total number of red marbles that Toby and Ling obtain.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial in $t$.
\item Use the probability generating function of $Z$ to find $\operatorname { Var } ( Z )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q3 [10]}}