| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Given PGF manipulation and properties |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question covering PGF manipulation with multiple techniques: finding constants using G(1)=1, extracting probabilities via differentiation, transforming PGFs for linear combinations, combining independent variables, and using calculus to find variance. While each individual step is standard for FS1, the multi-part nature requiring fluency across several PGF properties and the calculus-based variance calculation (requiring G''(1) and G'(1)) makes this moderately challenging, though still within expected FS1 scope. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G_X(1) = 1\) | M1 | 2.1 |
| \(k \times 3^5 = 1\) \(\therefore k = \dfrac{1}{243}\) | A1*cso | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=2)\) is coefficient of \(t^2\) so \(G_X(t) = k\left(...+{}^5C_2(2t)^2+...\right)\) | M1 | 1.1b |
| \(P(X=2) = \dfrac{40}{243}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G_W(t) = \dfrac{t^3}{243}\left(1+2(t^2)\right)^5\) | M1 | 3.1a |
| \(G_W(t) = \dfrac{t^3}{243}(1+2t^2)^5\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G_U(t) = \dfrac{1}{243}(1+2t)^5 \times \dfrac{t(1+2t)^2}{9}\) | M1 | 3.1a |
| \(= \dfrac{t(1+2t)^7}{2187}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G_U'(t) = \dfrac{14t(1+2t)^6}{2187} + \dfrac{(1+2t)^7}{2187}\) | M1 | 2.1 |
| \(G_U'(1) = \dfrac{17}{3}\) | A1ft | 1.1b |
| \(G_U''(t) = \dfrac{168t(1+2t)^5}{2187} + \dfrac{14(1+2t)^6}{2187} + \dfrac{14(1+2t)^6}{2187}\) | M1 | 2.1 |
| \(G_U''(1) = 28\) | A1 | 1.1b |
| \(\text{Var}(U) = 28 + \dfrac{17}{3} - \left(\dfrac{17}{3}\right)^2\) | M1 | 2.1 |
| \(= \dfrac{14}{9}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G_X''(t) = A(1+2t)^3\) | M1 | |
| \(G_X'(1) = \dfrac{10}{3}\) and \(G_X''(1) = \dfrac{80}{9}\) | A1ft | |
| \(G_Y''(t) = H(8+24t)\) | M1 | |
| \(G_Y'(1) = \dfrac{7}{3}\) and \(G_Y''(1) = \dfrac{32}{9}\) | A1 | |
| Using \(G_U''(1) + G_U'(1) - (G_U'(1))^2\) to find \(\text{Var}(X)\), \(\text{Var}(Y)\) and \(\text{Var}(U)\) | M1 | |
| \(\dfrac{14}{9}\) or awrt 1.56 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Stating \(G_X(1) = 1\), e.g. \(G_X(1) = k(1+2)^5 = 1\), so \(k(1+2)^5 = 1\) | M1 | Allow verification: \(\frac{1}{243} \times 3^5 = 1\) |
| Fully correct proof with no errors, substituting \(t=1\), verification needs \(G_X(1) = 1\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempting to find the coefficient of \(t^2\) | M1 | |
| \(\frac{40}{243}\) or awrt 0.165 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Realising the need to multiply through by \(t^3\) or substitute \(t^2\) for \(t\) | M1 | |
| \(\frac{t^3}{243}(1+2t^2)^5\) oe, e.g. \(\frac{t^3}{243}(1+10t^2+40t^4+80t^6+80t^8+32t^{10})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Realising the need to use \(G_U(t) = G_X(t) \times G_Y(t)\) | M1 | |
| \(\frac{t(1+2t)^7}{2187}\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to differentiate \(G(u)\), e.g. \(G_U'(t) = At(1+2t)^6 + B(1+2t)^7\), from part (d) in form \(kt(1+2t)^n\) where \(n \geq 5\) | M1 | |
| \(\frac{17}{3}\) or awrt 5.67 | A1ft | |
| Attempting second derivative, e.g. \(G_U''(t) = Ct(1+2t)^5 + D(1+2t)^6\), from part (d) in form \(kt(1+2t)^n\) where \(n \geq 5\) | M1 | |
| 28 | A1 | |
| Using \(G_U''(1) + G_U'(1) - \left(G_U'(1)\right)^2\) with their values | M1 | |
| \(\frac{14}{9}\) or awrt 1.56 | A1 |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G_X(1) = 1$ | M1 | 2.1 |
| $k \times 3^5 = 1$ $\therefore k = \dfrac{1}{243}$ | A1*cso | 1.1b |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=2)$ is coefficient of $t^2$ so $G_X(t) = k\left(...+{}^5C_2(2t)^2+...\right)$ | M1 | 1.1b |
| $P(X=2) = \dfrac{40}{243}$ | A1 | 1.1b |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G_W(t) = \dfrac{t^3}{243}\left(1+2(t^2)\right)^5$ | M1 | 3.1a |
| $G_W(t) = \dfrac{t^3}{243}(1+2t^2)^5$ | A1 | 1.1b |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G_U(t) = \dfrac{1}{243}(1+2t)^5 \times \dfrac{t(1+2t)^2}{9}$ | M1 | 3.1a |
| $= \dfrac{t(1+2t)^7}{2187}$ | A1 | 1.1b |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G_U'(t) = \dfrac{14t(1+2t)^6}{2187} + \dfrac{(1+2t)^7}{2187}$ | M1 | 2.1 |
| $G_U'(1) = \dfrac{17}{3}$ | A1ft | 1.1b |
| $G_U''(t) = \dfrac{168t(1+2t)^5}{2187} + \dfrac{14(1+2t)^6}{2187} + \dfrac{14(1+2t)^6}{2187}$ | M1 | 2.1 |
| $G_U''(1) = 28$ | A1 | 1.1b |
| $\text{Var}(U) = 28 + \dfrac{17}{3} - \left(\dfrac{17}{3}\right)^2$ | M1 | 2.1 |
| $= \dfrac{14}{9}$ | A1 | 1.1b |
## Part (e) ALT:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G_X''(t) = A(1+2t)^3$ | M1 | |
| $G_X'(1) = \dfrac{10}{3}$ and $G_X''(1) = \dfrac{80}{9}$ | A1ft | |
| $G_Y''(t) = H(8+24t)$ | M1 | |
| $G_Y'(1) = \dfrac{7}{3}$ and $G_Y''(1) = \dfrac{32}{9}$ | A1 | |
| Using $G_U''(1) + G_U'(1) - (G_U'(1))^2$ to find $\text{Var}(X)$, $\text{Var}(Y)$ and $\text{Var}(U)$ | M1 | |
| $\dfrac{14}{9}$ or awrt 1.56 | A1 | |
# Question (Previous - Probability Generating Functions):
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Stating $G_X(1) = 1$, e.g. $G_X(1) = k(1+2)^5 = 1$, so $k(1+2)^5 = 1$ | M1 | Allow verification: $\frac{1}{243} \times 3^5 = 1$ |
| Fully correct proof with no errors, substituting $t=1$, verification needs $G_X(1) = 1$ | A1* | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempting to find the coefficient of $t^2$ | M1 | |
| $\frac{40}{243}$ or awrt 0.165 | A1 | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Realising the need to multiply through by $t^3$ or substitute $t^2$ for $t$ | M1 | |
| $\frac{t^3}{243}(1+2t^2)^5$ oe, e.g. $\frac{t^3}{243}(1+10t^2+40t^4+80t^6+80t^8+32t^{10})$ | A1 | |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Realising the need to use $G_U(t) = G_X(t) \times G_Y(t)$ | M1 | |
| $\frac{t(1+2t)^7}{2187}$ oe | A1 | |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to differentiate $G(u)$, e.g. $G_U'(t) = At(1+2t)^6 + B(1+2t)^7$, from part (d) in form $kt(1+2t)^n$ where $n \geq 5$ | M1 | |
| $\frac{17}{3}$ or awrt 5.67 | A1ft | |
| Attempting second derivative, e.g. $G_U''(t) = Ct(1+2t)^5 + D(1+2t)^6$, from part (d) in form $kt(1+2t)^n$ where $n \geq 5$ | M1 | |
| 28 | A1 | |
| Using $G_U''(1) + G_U'(1) - \left(G_U'(1)\right)^2$ with their values | M1 | |
| $\frac{14}{9}$ or awrt 1.56 | A1 | |
---\begin{enumerate}
\item The probability generating function of the random variable $X$ is
\end{enumerate}
$$\mathrm { G } _ { X } ( t ) = k ( 1 + 2 t ) ^ { 5 }$$
where $k$ is a constant.\\
(a) Show that $k = \frac { 1 } { 243 }$\\
(b) Find $\mathrm { P } ( X = 2 )$\\
(c) Find the probability generating function of $W = 2 X + 3$
The probability generating function of the random variable $Y$ is
$$\mathrm { G } _ { Y } ( t ) = \frac { t ( 1 + 2 t ) ^ { 2 } } { 9 }$$
Given that $X$ and $Y$ are independent,\\
(d) find the probability generating function of $U = X + Y$ in its simplest form.\\
(e) Use calculus to find the value of $\operatorname { Var } ( U )$
\hfill \mbox{\textit{Edexcel FS1 2021 Q6 [14]}}