| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| 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 construction of PGFs from first principles, finding unknown parameters from PGF coefficients, using the independence property (G_Z = G_X × G_Y), and extracting expectations from PGFs. While the individual probability calculations are straightforward (hypergeometric for part a, independent Bernoulli trials for part b), the PGF framework and multi-step nature elevate this above standard A-level, though it remains a fairly routine Further Maths question testing core PGF techniques. |
| Spec | 5.02a Discrete probability distributions: general |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(BB) = \frac{6}{10}\times\frac{5}{9} = \frac{1}{3}\), \(P(RB, BR) = \frac{6}{10}\times\frac{4}{9} = \frac{8}{15}\), \(P(RR) = \frac{4}{10}\times\frac{3}{9} = \frac{2}{15}\) | M1 | |
| \(\frac{1}{3} + \frac{8}{15}t + \frac{2}{15}t^2\) | A1 FT | FT *their* probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(1\ H) = \frac{2}{3}(1-p) + \frac{1}{3}p = \frac{7}{12}\) | M1 | |
| \(p = \frac{1}{4}\) | A1 | |
| PGF of \(Y = \frac{1}{4} + \frac{7}{12}t + \frac{1}{6}t^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PGF of \(Z = \left(\frac{1}{3} + \frac{8}{15}t + \frac{2}{15}t^2\right)\!\left(\frac{1}{4} + \frac{7}{12}t + \frac{1}{6}t^2\right)\) | B1M1 | |
| \(= \frac{1}{180}\!\left(15 + 59t + 72t^2 + 30t^3 + 4t^4\right)\) | A1 | AEF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to differentiate their \(G_Z(t)\) and evaluate \(G'_Z(1)\) | M1 | |
| \(E(Z) = (59 + 144 + 90 + 16)/180 = \dfrac{309}{180} = 1.72\) | A1 |
## Question 6:
### Part 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(BB) = \frac{6}{10}\times\frac{5}{9} = \frac{1}{3}$, $P(RB, BR) = \frac{6}{10}\times\frac{4}{9} = \frac{8}{15}$, $P(RR) = \frac{4}{10}\times\frac{3}{9} = \frac{2}{15}$ | M1 | |
| $\frac{1}{3} + \frac{8}{15}t + \frac{2}{15}t^2$ | A1 FT | FT *their* probabilities |
### Part 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(1\ H) = \frac{2}{3}(1-p) + \frac{1}{3}p = \frac{7}{12}$ | M1 | |
| $p = \frac{1}{4}$ | A1 | |
| PGF of $Y = \frac{1}{4} + \frac{7}{12}t + \frac{1}{6}t^2$ | A1 | |
### Part 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| PGF of $Z = \left(\frac{1}{3} + \frac{8}{15}t + \frac{2}{15}t^2\right)\!\left(\frac{1}{4} + \frac{7}{12}t + \frac{1}{6}t^2\right)$ | B1M1 | |
| $= \frac{1}{180}\!\left(15 + 59t + 72t^2 + 30t^3 + 4t^4\right)$ | A1 | AEF |
### Part 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to differentiate their $G_Z(t)$ and evaluate $G'_Z(1)$ | M1 | |
| $E(Z) = (59 + 144 + 90 + 16)/180 = \dfrac{309}{180} = 1.72$ | A1 | |6 A bag contains 4 red balls and 6 blue balls. Rassa selects two balls at random, without replacement, from the bag. The number of red balls selected by Rassa is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function, $\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )$, of $X$.\\
Rassa also tosses two coins. One coin is biased so that the probability of a head is $\frac { 2 } { 3 }$. The other coin is biased so that the probability of a head is $p$. The probability generating function of $Y$, the number of heads obtained by Rassa, is $\mathrm { G } _ { Y } ( \mathrm { t } )$. The coefficient of $t$ in $\mathrm { G } _ { Y } ( \mathrm { t } )$ is $\frac { 7 } { 12 }$.
\item Find $\mathrm { G } _ { Y } ( \mathrm { t } )$.\\
The random variable $Z$ is the sum of the number of red balls selected and the number of heads obtained by Rassa.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.
\item Use the probability generating function of $Z$ to find $\mathrm { E } ( Z )$.\\
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 2020 Q6 [10]}}