| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find component PGF from sum PGF |
| Difficulty | Challenging +1.2 This question requires understanding that G_Y(t) = [G_X(t)]² for Y = X₁ + X₂, then finding G_X(t) by taking the square root of a polynomial. Part (a) is routine application of PGF formulas for variance. Part (b) requires recognizing the relationship and performing polynomial square root extraction, which is moderately challenging but follows a standard Further Maths technique with no novel insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Y'(t) = 0.2 + 0.74t + 0.9t^2 + 0.36t^3\); \(G_Y''(t) = 0.74 + 1.8t + 1.08t^2\) | M1 | PGF differentiated twice |
| \(G_Y'(1) = 2.2\); \(G_Y''(1) = 3.62\) | A1 | |
| \(\text{Var}(Y) = G_Y''(1) + G_Y'(1) - (G_Y'(1))^2 = 3.62 + 2.2 - 2.2^2\) | M1 | Correct formula used and attempt at Var\((Y)\) |
| \(0.98\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G_Y(t) = \left(a + bt + ct^2\right)^2\) | M1* | Correct method |
| \(a^2 = 0.04\), \(a = 0.2\); \(c^2 = 0.09\), \(c = 0.3\); \(2ab = 0.2\), \(b = \frac{0.1}{0.2} = 0.5\) | depM1 | Attempt to find all coefficients |
| \(G_X(t) = 0.2 + 0.5t + 0.3t^2\) | A1 | Note: \(G_Y = \frac{1}{100}(1+t)^2(2+3t)^2\) M1, accept without \(\frac{1}{100}\). So \(G_X = \frac{1}{10}(1+t)(2+3t)\) dep M1, must have \(\frac{1}{10}\). Final answer A1 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Y'(t) = 0.2 + 0.74t + 0.9t^2 + 0.36t^3$; $G_Y''(t) = 0.74 + 1.8t + 1.08t^2$ | M1 | PGF differentiated twice |
| $G_Y'(1) = 2.2$; $G_Y''(1) = 3.62$ | A1 | |
| $\text{Var}(Y) = G_Y''(1) + G_Y'(1) - (G_Y'(1))^2 = 3.62 + 2.2 - 2.2^2$ | M1 | Correct formula used and attempt at Var$(Y)$ |
| $0.98$ | A1 | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Y(t) = \left(a + bt + ct^2\right)^2$ | M1* | Correct method |
| $a^2 = 0.04$, $a = 0.2$; $c^2 = 0.09$, $c = 0.3$; $2ab = 0.2$, $b = \frac{0.1}{0.2} = 0.5$ | depM1 | Attempt to find all coefficients |
| $G_X(t) = 0.2 + 0.5t + 0.3t^2$ | A1 | Note: $G_Y = \frac{1}{100}(1+t)^2(2+3t)^2$ M1, accept without $\frac{1}{100}$. So $G_X = \frac{1}{10}(1+t)(2+3t)$ dep M1, must have $\frac{1}{10}$. Final answer A1 |2 The probability generating function, $\mathrm { G } _ { Y } ( t )$, of the random variable $Y$ is given by
$$G _ { Y } ( t ) = 0.04 + 0.2 t + 0.37 t ^ { 2 } + 0.3 t ^ { 3 } + 0.09 t ^ { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\operatorname { Var } ( Y )$.\\
The random variable $Y$ is the sum of two independent observations of the random variable $X$.
\item Find the probability generating function of $X$, giving your answer as a polynomial in $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q2 [7]}}