| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Determine constant in PGF |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question on PGFs requiring multiple standard techniques: using G(1)=1 to find c, differentiation for E(X), combining independent PGFs, and coefficient extraction. While methodical, it's routine application of PGF theory without novel insight, placing it moderately above average difficulty. |
| 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 | Mark | Guidance |
| \(c = \frac{1}{32}\) | B1 | Accept 0.03125 or 0.0313. |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G_X(t) = ct(1+t)^5\), \(G'_X(t) = c\left(5t(1+t)^4 + (1+t)^5\right)\) | M1 | Must use product rule OR differentiate their expansion of \(G\). Allow one error in their differentiation. Condone missing \(c\). |
| \(E(X) = G'_X(1) = c(80 + 32) = 3.5\) | A1 | Correct work only. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G_Y(t) = c^2t^2(1+t)^{10}\) | *M1 | Or \(c^2\left(t^2 + 10t^3 + 45t^4 + 120t^5 + 210t^6 + 252t^7 + 210t^8 + 120t^9 + 45t^{10} + 10t^{11} + t^{12}\right)\) |
| \(G'_Y(t) = c^2\left(10t^2(1+t)^9 + 2t(1+t)^{10}\right)\); \(G''_Y(t) = c^2\left(90t^2(1+t)^8 + 40t(1+t)^9 + 2(1+t)^{10}\right)\) | DM1 | Differentiate their \(G_Y(t)\) twice, allow one error. |
| \(G'_Y(1) = 7\), \(G''_Y(1) = 44.5\); \(\text{Var}(Y) = G''_Y(1) + G'_Y(1) - \left(G'_Y(1)\right)^2 = 2.5\) | M1 A1 | Use correct formula and substitute values for \(t=1\) (may be in terms of \(c\)). A1 for correct work only. |
| Alternative: \(\text{Var}(Y) = 2\text{Var}(X)\) | *M1 | |
| Differentiate \(G_X(t)\) twice | DM1 | Note that \(G'_X(t)\) has already been found in part (b). |
| \(\text{Var}(X) = G''_X(1) + G'_X(1) - \left(G'_X(1)\right)^2\) | M1 | Use correct formula and substitute values for \(t=1\). |
| \(1.25 \times 2 = 2.5\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y=5)\) = coefficient of \(t^5\) in expansion of \(c^2t^2(1+t)^{10}\); Required term is \(c^2 \times 120\); Expansion \(= c^2t^2(1+10t+45t^2+120t^3+\ldots)\) | M1 | Identify required term. |
| \(\frac{15}{128}\) | A1 | \(0.117\), \(\frac{120}{1024}\) or any equivalent fraction. |
# Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $c = \frac{1}{32}$ | B1 | Accept 0.03125 or 0.0313. |
| **Total: 1** | | |
---
# Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G_X(t) = ct(1+t)^5$, $G'_X(t) = c\left(5t(1+t)^4 + (1+t)^5\right)$ | M1 | Must use product rule OR differentiate their expansion of $G$. Allow one error in their differentiation. Condone missing $c$. |
| $E(X) = G'_X(1) = c(80 + 32) = 3.5$ | A1 | Correct work only. |
| **Total: 2** | | |
---
# Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G_Y(t) = c^2t^2(1+t)^{10}$ | *M1 | Or $c^2\left(t^2 + 10t^3 + 45t^4 + 120t^5 + 210t^6 + 252t^7 + 210t^8 + 120t^9 + 45t^{10} + 10t^{11} + t^{12}\right)$ |
| $G'_Y(t) = c^2\left(10t^2(1+t)^9 + 2t(1+t)^{10}\right)$; $G''_Y(t) = c^2\left(90t^2(1+t)^8 + 40t(1+t)^9 + 2(1+t)^{10}\right)$ | DM1 | Differentiate their $G_Y(t)$ twice, allow one error. |
| $G'_Y(1) = 7$, $G''_Y(1) = 44.5$; $\text{Var}(Y) = G''_Y(1) + G'_Y(1) - \left(G'_Y(1)\right)^2 = 2.5$ | M1 A1 | Use correct formula and substitute values for $t=1$ (may be in terms of $c$). A1 for correct work only. |
| **Alternative:** $\text{Var}(Y) = 2\text{Var}(X)$ | *M1 | |
| Differentiate $G_X(t)$ twice | DM1 | Note that $G'_X(t)$ has already been found in part (b). |
| $\text{Var}(X) = G''_X(1) + G'_X(1) - \left(G'_X(1)\right)^2$ | M1 | Use correct formula and substitute values for $t=1$. |
| $1.25 \times 2 = 2.5$ | A1 | |
| **Total: 4** | | |
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y=5)$ = coefficient of $t^5$ in expansion of $c^2t^2(1+t)^{10}$; Required term is $c^2 \times 120$; Expansion $= c^2t^2(1+10t+45t^2+120t^3+\ldots)$ | M1 | Identify required term. |
| $\frac{15}{128}$ | A1 | $0.117$, $\frac{120}{1024}$ or any equivalent fraction. |
---4 The random variable $X$ has probability generating function $\mathrm { G } _ { X } ( t )$ given by
$$\mathrm { G } _ { X } ( t ) = \operatorname { ct } ( 1 + t ) ^ { 5 }$$
where $c$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $c$.
\item Find the value of $\mathrm { E } ( X )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-06_2718_33_141_2014}
The random variable $Y$ is the sum of two independent values of $X$.
\item Write down the probability generating function of $Y$ and hence find $\operatorname { Var } ( Y )$.
\item Find $\mathrm { P } ( Y = 5 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q4 [9]}}