| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Standard +0.8 This is a multi-part Further Maths statistics question requiring: (a) finding a PGF from a hypergeometric distribution via enumeration, (b) finding a PGF for a binomial distribution, (c) using the independence property G_Z(t) = G_X(t)G_Y(t), and (d) using derivatives of PGFs to find mean and variance. While systematic, it requires knowledge of PGF theory, careful probability calculations without replacement, and algebraic manipulation across multiple parts—moderately challenging for Further Maths. |
| Spec | 5.02a Discrete probability distributions: general |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(3R,\ 2R,\ 1R,\ 0R) = \frac{1}{5}, \frac{3}{5}, \frac{1}{5}, 0\) | M1 | At least 2 correct probabilities used in a polynomial |
| \(G_X(t) = \frac{1}{5}t + \frac{3}{5}t^2 + \frac{1}{5}t^3\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2H,\ 1H,\ 0H) = \frac{1}{16}, \frac{6}{16}, \frac{9}{16}\) | M1 | One correct probability and all three adding to 1 used in a polynomial |
| \(G_Y(t) = \frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to multiply results from part (a) and part (b) | M1 | |
| Obtain polynomial | M1 | |
| \(\frac{9}{80}t + \frac{33}{80}t^2 + \frac{28}{80}t^3 + \frac{9}{80}t^4 + \frac{1}{80}t^5\) | A1 | Accept exact equivalent decimals |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Z) = G_Z'(1) = \frac{1}{80}(9 + 66 + 84 + 36 + 5)\) | M1 | Differentiate and put \(t = 1\) |
| \(= \frac{200}{80} = 2.5\) | A1 | |
| \(G_Z''(1) = \frac{1}{80}(66 + 168 + 108 + 20) = 4.525 \left[= \frac{362}{80}\right]\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(Z) = G''(1) + G'(1) - (G'(1))^2\) | M1 | Use result. |
| \(\dfrac{362}{80} + \dfrac{200}{80} - \left(\dfrac{200}{80}\right)^2 = 0.775 = \left[\dfrac{31}{40}\right]\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(3R,\ 2R,\ 1R,\ 0R) = \frac{1}{5}, \frac{3}{5}, \frac{1}{5}, 0$ | M1 | At least 2 correct probabilities used in a polynomial |
| $G_X(t) = \frac{1}{5}t + \frac{3}{5}t^2 + \frac{1}{5}t^3$ | A1 | |
| **Total** | **2** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2H,\ 1H,\ 0H) = \frac{1}{16}, \frac{6}{16}, \frac{9}{16}$ | M1 | One correct probability and all three adding to 1 used in a polynomial |
| $G_Y(t) = \frac{9}{16} + \frac{6}{16}t + \frac{1}{16}t^2$ | A1 | |
| **Total** | **2** | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to multiply results from part (a) and part (b) | M1 | |
| Obtain polynomial | M1 | |
| $\frac{9}{80}t + \frac{33}{80}t^2 + \frac{28}{80}t^3 + \frac{9}{80}t^4 + \frac{1}{80}t^5$ | A1 | Accept exact equivalent decimals |
| **Total** | **3** | |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Z) = G_Z'(1) = \frac{1}{80}(9 + 66 + 84 + 36 + 5)$ | M1 | Differentiate and put $t = 1$ |
| $= \frac{200}{80} = 2.5$ | A1 | |
| $G_Z''(1) = \frac{1}{80}(66 + 168 + 108 + 20) = 4.525 \left[= \frac{362}{80}\right]$ | M1 | |
## Question 6(d):
$\text{Var}(Z) = G''(1) + G'(1) - (G'(1))^2$ | M1 | Use result.
$\dfrac{362}{80} + \dfrac{200}{80} - \left(\dfrac{200}{80}\right)^2 = 0.775 = \left[\dfrac{31}{40}\right]$ | A1 |
**Total: 5**6 Tanji has a bag containing 4 red balls and 2 blue balls. He selects 3 balls at random from the bag, without replacement. The number of red balls selected by Tanji is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function $\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )$ of $X$.\\
Tanji also has two coins, each biased so that the probability of obtaining a head when it is thrown is $\frac { 1 } { 4 }$. He throws the two coins at the same time. The number of heads obtained is denoted by $Y$.
\item Find the probability generating function $\mathrm { G } _ { Y } ( \mathrm { t } )$ of $Y$.\\
The random variable $Z$ is the sum of the number of red balls selected by Tanji and the number of heads obtained.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.
\item Use the probability generating function of $Z$ to find $E ( Z )$ and $\operatorname { Var } ( Z )$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q6 [12]}}