| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find component PGF from sum PGF |
| Difficulty | Challenging +1.2 This is a straightforward application of PGF properties: (a) uses the standard derivative formula for E(Y), (b) requires recognizing that G_Y(t) = [G_X(t)]^2 so G_X(t) is the square root, and (c) involves expanding the PGF and reading off a coefficient. While it requires knowledge of PGF theory (a Further Maths topic), the execution is mechanical with no novel problem-solving required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Y(t) = \dfrac{t^2}{(4-3t)^4}\), \(G'_Y(t) = \dfrac{(4-3t)^4(2t) - (-12t^2)(4-3t)^3}{(4-3t)^8}\) or \(\dfrac{2t}{(4-3t)^4} + \dfrac{12t^2}{(4-3t)^5}\) | M1 | 'Two' terms obtained, correct denominator(s) |
| \(E(Y) = G'_Y(1) = 14\) | M1 A1 | \(t = 1\) in their expression. CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_X(t) = \dfrac{t}{(4-3t)^2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_X(t) = t(4-3t)^{-2} = \dfrac{t}{16}\left(1 - \dfrac{3}{4}t\right)^{-2} = \dfrac{t}{16}\left(1 + \dfrac{3}{2}t - 3 \times \dfrac{9}{16}t^2 + 4 \times \dfrac{27}{64}t^3 + \ldots\right)\) | M1 | Expansion as far as \(t^3\) OR term in \(t^4\) calculated |
| Use \(P(X=4)\) = their coefficient of \(t^4\) | M1 | Find the numerical value of their coefficient of \(t^4\) |
| \(\dfrac{27}{256}\) | A1 | Accept \(0.105\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Y(t) = \dfrac{t^2}{(4-3t)^4}$, $G'_Y(t) = \dfrac{(4-3t)^4(2t) - (-12t^2)(4-3t)^3}{(4-3t)^8}$ or $\dfrac{2t}{(4-3t)^4} + \dfrac{12t^2}{(4-3t)^5}$ | M1 | 'Two' terms obtained, correct denominator(s) |
| $E(Y) = G'_Y(1) = 14$ | M1 A1 | $t = 1$ in their expression. CWO |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_X(t) = \dfrac{t}{(4-3t)^2}$ | B1 | |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_X(t) = t(4-3t)^{-2} = \dfrac{t}{16}\left(1 - \dfrac{3}{4}t\right)^{-2} = \dfrac{t}{16}\left(1 + \dfrac{3}{2}t - 3 \times \dfrac{9}{16}t^2 + 4 \times \dfrac{27}{64}t^3 + \ldots\right)$ | M1 | Expansion as far as $t^3$ OR term in $t^4$ calculated |
| Use $P(X=4)$ = their coefficient of $t^4$ | M1 | Find the numerical value of their coefficient of $t^4$ |
| $\dfrac{27}{256}$ | A1 | Accept $0.105$ |
---4 The random variable $Y$ is the sum of two independent observations of the random variable $X$. The probability generating function $\mathrm { G } _ { Y } ( \mathrm { t } )$ of $Y$ is given by
$$G _ { Y } ( t ) = \frac { t ^ { 2 } } { ( 4 - 3 t ) ^ { 4 } }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( \mathrm { Y } )$.
\item Write down an expression for the probability generating function of $X$.
\item Find $\mathrm { P } ( X = 4 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q4 [7]}}