| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Multiple independent coins/dice |
| Difficulty | Standard +0.3 This is a straightforward application of PGF theory for independent random variables. Part (a) requires constructing a binomial PGF using the standard formula (q+pt)^n. Part (b) uses the key property that PGFs multiply for independent sums. Part (c) applies the standard result G'(1)=E(Z). All steps are direct applications of bookwork with minimal problem-solving required, making this slightly easier than average. |
| 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 | Marks | Guidance |
| \(P(0\ H) = \frac{1}{3}\times\frac{1}{3}\times\frac{1}{3} = \frac{1}{27}\); \(P(1\ H) = \frac{2}{3}\times\frac{1}{3}\times\frac{1}{3}\times 3 = \frac{6}{27}\); \(P(2\ H) = \frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}\times 3 = \frac{12}{27}\); \(P(3\ H) = \frac{2}{3}\times\frac{2}{3}\times\frac{2}{3} = \frac{8}{27}\) | M1 | At least 2 probabilities correct |
| \(G_X(t) = \frac{1}{27} + \frac{6}{27}t + \frac{12}{27}t^2 + \frac{8}{27}t^3\) | M1A1 | Correct form, ft their probabilities |
| OR: PGF for one coin is \(\frac{1}{3} + \frac{2}{3}t\) | M1 | |
| For 3 coins, PGF is \(\left(\frac{1}{3}+\frac{2}{3}t\right)^3\) | M1A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Z(t) = \frac{1}{108}(1+2t+t^2)(1+6t+12t^2+8t^3)\) | M1 | Correct method |
| \(\frac{1}{108}(1+8t+25t^2+38t^3+28t^4+8t^5)\) | M1 | Obtain quintic polynomial |
| \(\frac{1}{108}+\frac{2}{27}t+\frac{25}{108}t^2+\frac{19}{54}t^3+\frac{7}{27}t^4+\frac{2}{27}t^5\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Z'(t) = \frac{1}{108}(8+50t+114t^2+112t^3+40t^4)\); \(E(Z) = \frac{1}{108}(8+50+114+112+40)\) | M1 | Differentiate and evaluate at \(t=1\) |
| \(\frac{324}{108} = 3\) | A1 | |
| Alternative: \(\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{1}{2}+\frac{1}{2}\) | M1 | Summing expected values for each coin |
| \(3\) | A1 | |
| 2 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0\ H) = \frac{1}{3}\times\frac{1}{3}\times\frac{1}{3} = \frac{1}{27}$; $P(1\ H) = \frac{2}{3}\times\frac{1}{3}\times\frac{1}{3}\times 3 = \frac{6}{27}$; $P(2\ H) = \frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}\times 3 = \frac{12}{27}$; $P(3\ H) = \frac{2}{3}\times\frac{2}{3}\times\frac{2}{3} = \frac{8}{27}$ | M1 | At least 2 probabilities correct |
| $G_X(t) = \frac{1}{27} + \frac{6}{27}t + \frac{12}{27}t^2 + \frac{8}{27}t^3$ | M1A1 | Correct form, ft their probabilities |
| OR: PGF for one coin is $\frac{1}{3} + \frac{2}{3}t$ | M1 | |
| For 3 coins, PGF is $\left(\frac{1}{3}+\frac{2}{3}t\right)^3$ | M1A1 | |
| | **3** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z(t) = \frac{1}{108}(1+2t+t^2)(1+6t+12t^2+8t^3)$ | M1 | Correct method |
| $\frac{1}{108}(1+8t+25t^2+38t^3+28t^4+8t^5)$ | M1 | Obtain quintic polynomial |
| $\frac{1}{108}+\frac{2}{27}t+\frac{25}{108}t^2+\frac{19}{54}t^3+\frac{7}{27}t^4+\frac{2}{27}t^5$ | A1 | |
| | **3** | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z'(t) = \frac{1}{108}(8+50t+114t^2+112t^3+40t^4)$; $E(Z) = \frac{1}{108}(8+50+114+112+40)$ | M1 | Differentiate and evaluate at $t=1$ |
| $\frac{324}{108} = 3$ | A1 | |
| **Alternative:** $\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\frac{1}{2}+\frac{1}{2}$ | M1 | Summing expected values for each coin |
| $3$ | A1 | |
| | **2** | |4 Jason has three biased coins. For each coin the probability of obtaining a head when it is thrown is $\frac { 2 } { 3 }$. Jason throws all three coins. The number of heads obtained is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function $\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )$ of $X$.\\
Jason also has two unbiased coins. He throws all five coins. The number of heads obtained from the two unbiased coins is denoted by $Y$. It is given that $G _ { Y } ( t ) = \frac { 1 } { 4 } + \frac { 1 } { 2 } t + \frac { 1 } { 4 } t ^ { 2 }$. The random variable $Z$ is the total number of heads obtained when Jason throws all five coins.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.
\item Find $\mathrm { E } ( \mathrm { Z } )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q4 [8]}}