| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| 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 systematic application of PGF theory. Part (a) involves multiplying PGFs for independent Bernoulli trials with different probabilities. Part (b) requires finding the PGF for binomial dice rolls and multiplying with part (a). Part (c) uses G'(1) to find expectation. While mechanical and requiring careful algebra, it's a textbook application of PGF techniques without novel insight—typical for Further Maths but above standard A-level. |
| Spec | 5.02a Discrete probability distributions: general |
| \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_444_33_106_2013} | \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_443_33_675_2013} | \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_440_33_1247_2013} | \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_441_33_1816_2013} | \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_443_31_2385_2015} |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(0 \text{ heads}) = \frac{8}{30}\), \(P(1 \text{ head}) = \frac{14}{30}\), \(P(2 \text{ heads}) = \frac{7}{30}\), \(P(3 \text{ heads}) = \frac{1}{30}\) | B1 | All correct |
| \(G_X(t) = \frac{8}{30} + \frac{14}{30}t + \frac{7}{30}t^2 + \frac{1}{30}t^3\) | M1 | Cubic polynomial |
| A1 FT | FT their probabilities that sum to one |
| Answer | Marks | Guidance |
|---|---|---|
| \(G_Y(t) = \frac{25}{36} + \frac{10}{36}t + \frac{1}{36}t^2\) | B1 | |
| \(G_Z(t) = \frac{1}{30 \times 36}\left(\frac{8}{30} + \frac{14}{30}t + \frac{7}{30}t^2 + \frac{1}{30}t^3\right)(25 + 10t + t^2)\) | M1 | With attempt to multiply |
| \(\frac{1}{1080}(200 + 430t + 323t^2 + 109t^3 + 17t^4 + t^5)\) | M1 A1 | Obtain quintic polynomial |
| Answer | Marks | Guidance |
|---|---|---|
| \(G_Z{'}(1) = \frac{1}{1080}(430 + 646 + 327 + 68 + 5)\) | M1 | Differentiate and substitute \(t = 1\) |
| \(\frac{1476}{1080} = \frac{41}{30} = 1.37\) | A1 |
## Question 5(a):
| $P(0 \text{ heads}) = \frac{8}{30}$, $P(1 \text{ head}) = \frac{14}{30}$, $P(2 \text{ heads}) = \frac{7}{30}$, $P(3 \text{ heads}) = \frac{1}{30}$ | B1 | All correct |
|---|---|---|
| $G_X(t) = \frac{8}{30} + \frac{14}{30}t + \frac{7}{30}t^2 + \frac{1}{30}t^3$ | M1 | Cubic polynomial |
| | A1 FT | FT their probabilities that sum to one |
## Question 5(b):
| $G_Y(t) = \frac{25}{36} + \frac{10}{36}t + \frac{1}{36}t^2$ | B1 | |
|---|---|---|
| $G_Z(t) = \frac{1}{30 \times 36}\left(\frac{8}{30} + \frac{14}{30}t + \frac{7}{30}t^2 + \frac{1}{30}t^3\right)(25 + 10t + t^2)$ | M1 | With attempt to multiply |
| $\frac{1}{1080}(200 + 430t + 323t^2 + 109t^3 + 17t^4 + t^5)$ | M1 A1 | Obtain quintic polynomial |
## Question 5(c):
| $G_Z{'}(1) = \frac{1}{1080}(430 + 646 + 327 + 68 + 5)$ | M1 | Differentiate and substitute $t = 1$ |
|---|---|---|
| $\frac{1476}{1080} = \frac{41}{30} = 1.37$ | A1 | |
---5 Nikita has three coins. One coin is fair, one coin is biased so that the probability of obtaining a head is $\frac { 1 } { 3 }$ and the third coin is biased so that the probability of obtaining a head is $\frac { 1 } { 5 }$. The random variable $X$ is the number of heads that Nikita obtains when he throws all three coins at the same time.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function of $X$.\\
Rajesh has two fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6. The random variable $Y$ is the number of 4 s that Rajesh obtains when he throws the two dice.
The random variable $Z$ is the sum of the number of heads obtained by Nikita and the number of 4 s obtained by Rajesh.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.\\
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_444_33_106_2013}
& \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_443_33_675_2013}
& \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_440_33_1247_2013}
& \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_441_33_1816_2013}
& \includegraphics[max width=\textwidth, alt={}]{8b2a13d7-62f4-45a7-84c5-7d5bc870b8ce-10_443_31_2385_2015}
\\
\hline
\end{tabular}
\end{center}
\item Use your answer to part (b) to find $\mathrm { E } ( Z )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q5 [9]}}