| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find probabilities from PGF |
| Difficulty | Standard +0.8 This is part (b) of a PGF question requiring differentiation of a generating function four times and evaluation at zero, then dividing by 4!. While mechanically straightforward for Further Maths students familiar with PGFs, it requires careful algebraic manipulation and is more demanding than routine A-level questions. The multi-step differentiation and coefficient extraction places it moderately above average difficulty. |
| Spec | 5.02a Discrete probability distributions: general |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks |
|---|---|
| 4(a) | 4t |
| Answer | Marks | Guidance |
|---|---|---|
| 3−2t2 | M1 | kt |
| Answer | Marks | Guidance |
|---|---|---|
| so E(X) = G’(1) = 4 | A1 | CAO WWW |
| Answer | Marks | Guidance |
|---|---|---|
| 3−2t2 | M1 | OE. Differentiate, allow only numerical or sign slips. |
| Answer | Marks | Guidance |
|---|---|---|
| Var(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16 | M1 | Substitute their values into correct formula, dependent on |
| Answer | Marks | Guidance |
|---|---|---|
| [Var(X)] = 24 | A1 | CAO WWW |
| Answer | Marks |
|---|---|
| 4(b) | G ( t )= 1 = ( 3−2t2 )−1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 9 | M1 | Expand given expression or give expression for the term in t4 . |
| P(X = 4) = their coefficient of t4 | M1 | Found from legitimate method. |
| Answer | Marks | Guidance |
|---|---|---|
| 27 | A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | 4t
G’(t) =
( )2
3−2t2 | M1 | kt
Differentiate to obtain OE.
( )2
3−2t2
so E(X) = G’(1) = 4 | A1 | CAO WWW
12+24t2 ( )−2 ( )−3
G’’(t) = or 4 3−2t2 +32t2 3−2t2
( )3
3−2t2 | M1 | OE. Differentiate, allow only numerical or sign slips.
( ( ))2
Var(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16 | M1 | Substitute their values into correct formula, dependent on
attempt at G' '(t).
[Var(X)] = 24 | A1 | CAO WWW
5
--- 4(b) ---
4(b) | G ( t )= 1 = ( 3−2t2 )−1
X 3−2t2
1 2 4
= 1+ t2 + t4 +….
3 3 9 | M1 | Expand given expression or give expression for the term in t4 .
P(X = 4) = their coefficient of t4 | M1 | Found from legitimate method.
4
[P(X = 4) =]
27 | A1 | WWW
3
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find P$(X = 4)$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q4 [3]}}