| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Challenging +1.2 This is a multi-part Further Statistics question requiring probability distribution properties, conditional probability calculations, and variance transformations. Part (a) is straightforward expectation calculation. Part (b) requires systematic enumeration of game outcomes with two constraints (probabilities sum to 1, and P(Greg wins) = 1/6), leading to simultaneous equations. Part (c) uses standard variance transformation rules. While it involves several steps and careful bookkeeping, the techniques are all standard for FS1 with no novel insights required, making it moderately above average difficulty for A-level but routine for Further Maths students. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(w\) | 1 | 2 | 4 | 5 | \(c\) |
| \(\mathrm { P } ( W = w )\) | \(a + b\) | \(a\) | 0.3 | \(a\) | \(b\) |
| V349 SIHI NI IMIMM ION OC | VJYV SIHIL NI LIIIM ION OO | VJYV SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.5\) | B1 | \(\frac{5}{2}\) or \(2.5\) |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A complete strategy to find a value for \(a\) and a value for \(b\) | dM1 | Dependent on 3rd and 4th Method marks. Need 2 independent equations in \(a\) and \(b\), one equation must be prob \(= \frac{1}{6}\) and the other \(3a+2b=0.7\) oe, and an attempt to solve |
| Table listing 3 combinations for Greg to win: \((4,1), (4,2), (3,1)\) | M1 | For using contextual information to list 3 different combinations for Greg to win. Implied by \(4a+3b=1\) oe |
| \(\frac{1}{6}(2a+b)+\frac{1}{3}(a+b)=\frac{1}{6}\) oe | M1 | For using \(P(g)\times P(n)\) for each combination identified as a win for Greg \(=\frac{1}{6}\). Must be a linear equation in \(a\) and \(b\) with \(\frac{2}{3}\) terms on LHS, at least one correct and equal to \(\frac{1}{6}\) |
| \(4a+3b=1\) oe | ||
| \(3a+2b=0.7\) oe | M1 | For use of \(\sum P(W=w)=1\) |
| \(a=0.1\), \(b=0.2\) | A1 | For both values correct |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2E(W)-5=2.6\) | M1 | For translating the given mathematical context into an expression for \(E(W)\). May be implied by a correct equation or \(E(W)=3.8\) |
| \(E(W)=3.8\) | ||
| \(8a+b+cb+1.2[=3.8]\) or \("0.3"+2\times"0.1"+4\times0.3+5\times"0.1"+"0.2"c=["3.8"]\) | M1 | For use of \(\sum wP(W=w)[=3.8]\). If algebraic then at least 2 terms must be correct, if numerical at least 3 terms correct ft values of \(a\) and \(b\) |
| \(c=\frac{2.6-8"a"-"b"}{{"b"}}\) | ||
| \(c=8\) | A1 | cao |
| \(E(W^2)=30a+c^2b+b+4.8\) or \("0.3"+4\times"0.1"+16\times0.3+25\times"0.1"+"8"^2\times0.2\) | M1 | For use of \(\sum w^2P(W=w)\). If algebraic then at least 2 terms must be correct, if numerical at least 3 terms correct ft values of \(a\) and \(b\) |
| \(E(W^2)=20.8\) | ||
| \(\text{Var}(W)="20.8"-"3.8"^2\) or \(6.36\) | M1 | For use of \(\text{Var}(W)=E(W^2)-[E(W)]^2\) |
| \(\text{Var}(X)=25.44\) | A1ft | \(4\times\)"their \(\text{Var}(W)\)" ft their \(\text{Var}(W)\) provided \(a\), \(b\) and \(\text{Var}(W)>0\) and \(c>5\) |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([E(X)=]-3\times(a+b)-1\times a+0.9+5\times a+(2c-5)\times b\) | M1 | Allow with their \(a\), \(b\) and \(X\) values |
| \(-3\times(a+b)-1\times a+0.9+5\times a+(2c-5)\times b=2.6\) | M1 | Allow with their \(a\), \(b\) and \(X\) values |
| \(c=8\) | A1 | cao |
| Values of \(X\): \(-3,-1,3,5,2c-5\) | M1 | At least 3 correct |
| \(E(X^2)=9\times(a+b)+1\times a+2.7+25\times a+(2c-5)^2\times b = 32.2\) | M1 | Allow with their \(c\), \(a\), \(b\) and \(X\) values |
| \(\text{Var}(X)="32.2"-2.6^2 = 25.44\) | A1ft | ft their \(E(X^2)\) provided \(a\) and \(b>0\) and \(c>5\) |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.5$ | B1 | $\frac{5}{2}$ or $2.5$ |
| | **(1)** | |
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A complete strategy to find a value for $a$ and a value for $b$ | dM1 | Dependent on 3rd and 4th Method marks. Need 2 independent equations in $a$ and $b$, one equation must be prob $= \frac{1}{6}$ and the other $3a+2b=0.7$ oe, and an attempt to solve |
| Table listing 3 combinations for Greg to win: $(4,1), (4,2), (3,1)$ | M1 | For using contextual information to list 3 different combinations for Greg to win. Implied by $4a+3b=1$ oe |
| $\frac{1}{6}(2a+b)+\frac{1}{3}(a+b)=\frac{1}{6}$ oe | M1 | For using $P(g)\times P(n)$ for each combination identified as a win for Greg $=\frac{1}{6}$. Must be a linear equation in $a$ and $b$ with $\frac{2}{3}$ terms on LHS, at least one correct and equal to $\frac{1}{6}$ |
| $4a+3b=1$ oe | | |
| $3a+2b=0.7$ oe | M1 | For use of $\sum P(W=w)=1$ |
| $a=0.1$, $b=0.2$ | A1 | For both values correct |
| | **(5)** | |
---
## Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2E(W)-5=2.6$ | M1 | For translating the given mathematical context into an expression for $E(W)$. May be implied by a correct equation or $E(W)=3.8$ |
| $E(W)=3.8$ | | |
| $8a+b+cb+1.2[=3.8]$ or $"0.3"+2\times"0.1"+4\times0.3+5\times"0.1"+"0.2"c=["3.8"]$ | M1 | For use of $\sum wP(W=w)[=3.8]$. If algebraic then at least 2 terms must be correct, if numerical at least 3 terms correct ft values of $a$ and $b$ |
| $c=\frac{2.6-8"a"-"b"}{{"b"}}$ | | |
| $c=8$ | A1 | cao |
| $E(W^2)=30a+c^2b+b+4.8$ or $"0.3"+4\times"0.1"+16\times0.3+25\times"0.1"+"8"^2\times0.2$ | M1 | For use of $\sum w^2P(W=w)$. If algebraic then at least 2 terms must be correct, if numerical at least 3 terms correct ft values of $a$ and $b$ |
| $E(W^2)=20.8$ | | |
| $\text{Var}(W)="20.8"-"3.8"^2$ or $6.36$ | M1 | For use of $\text{Var}(W)=E(W^2)-[E(W)]^2$ |
| $\text{Var}(X)=25.44$ | A1ft | $4\times$"their $\text{Var}(W)$" ft their $\text{Var}(W)$ provided $a$, $b$ and $\text{Var}(W)>0$ and $c>5$ |
| | **(6)** | |
**Alternative for (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[E(X)=]-3\times(a+b)-1\times a+0.9+5\times a+(2c-5)\times b$ | M1 | Allow with their $a$, $b$ and $X$ values |
| $-3\times(a+b)-1\times a+0.9+5\times a+(2c-5)\times b=2.6$ | M1 | Allow with their $a$, $b$ and $X$ values |
| $c=8$ | A1 | cao |
| Values of $X$: $-3,-1,3,5,2c-5$ | M1 | At least 3 correct |
| $E(X^2)=9\times(a+b)+1\times a+2.7+25\times a+(2c-5)^2\times b = 32.2$ | M1 | Allow with their $c$, $a$, $b$ and $X$ values |
| $\text{Var}(X)="32.2"-2.6^2 = 25.44$ | A1ft | ft their $E(X^2)$ provided $a$ and $b>0$ and $c>5$ |
---\begin{enumerate}
\item A fair six-sided black die has faces numbered $1,2,2,3,3$ and 4
\end{enumerate}
The random variable $B$ represents the score when the black die is rolled.\\
(a) Write down the value of $\mathrm { E } ( B )$
A white die has 6 faces numbered $1,1,2,4,5$ and $c$ where $c > 5$\\
The discrete random variable $W$ represents the score when the white die is rolled and has probability distribution given by
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$w$ & 1 & 2 & 4 & 5 & $c$ \\
\hline
$\mathrm { P } ( W = w )$ & $a + b$ & $a$ & 0.3 & $a$ & $b$ \\
\hline
\end{tabular}
\end{center}
Greg and Nilaya play a game with these dice.\\
Greg throws the black die and Nilaya throws the white die. Greg wins the game if he scores at least two more than Nilaya, otherwise Greg loses.\\
The probability of Greg winning the game is $\frac { 1 } { 6 }$\\
(b) Find the value of $a$ and the value of $b$
Show your working clearly.
The random variable $X = 2 W - 5$\\
Given that $\mathrm { E } ( X ) = 2.6$\\
(c) find the exact value of $\operatorname { Var } ( X )$
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
V349 SIHI NI IMIMM ION OC & VJYV SIHIL NI LIIIM ION OO & VJYV SIHIL NI JIIYM ION OC \\
\hline
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel FS1 AS 2018 Q3 [12]}}