| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability concepts: independence of dice rolls, constructing a sample space, reading probabilities from equally likely outcomes, and calculating expectation and variance using standard formulas. All parts are routine applications of definitions with no problem-solving insight required, making it easier than average. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(t\) | 0 | 1 | 2 | 3 | 4 | 6 | 9 |
| \(\mathrm { P } ( T = t )\) | \(a\) | \(b\) | \(1 / 8\) | \(1 / 8\) | \(c\) | \(1 / 8\) | \(d\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(R=3 \cap B=0) = \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16}\) | M1, A1 | For \(\dfrac{1}{4} \times \dfrac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| All 0s correct in table | B1 | |
| All 1, 2, 3s correct in table | B1 | |
| All 4, 6, 9s correct in table | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = \dfrac{7}{16}\) | B1 | For \(\dfrac{7}{16}\) |
| \(b = c = d = \dfrac{1}{16}\) | B1, B1 | \(2^{\text{nd}}\) B1 for only one error in \(b,c,d\); \(3^{\text{rd}}\) B1 all of \(b,c,d = \dfrac{1}{16}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(T) = \left(1\times\dfrac{1}{16}\right)+\left(2\times\dfrac{1}{8}\right)+\left(3\times\dfrac{3}{8}\right)+\left(4\times\dfrac{1}{16}\right)+\ldots\) | M1 | For attempting \(\sum tP(T=t)\), 3 or more terms correct or correct ft. NB dividing by a number other than 1 scores M0 |
| \(= 2\dfrac{1}{4}\) or exact equivalent e.g. \(2.25\), \(\dfrac{9}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(T) = \left(1^2\times\dfrac{1}{16}\right)+\left(2^2\times\dfrac{1}{8}\right)+\left(3^2\times\dfrac{3}{8}\right)+\left(4^2\times\dfrac{1}{16}\right)+\ldots-\left(\dfrac{9}{4}\right)^2\) | M1A1, M1 | \(1^{\text{st}}\) M1 for attempt at \(E(T^2)\), 3+ terms correct; A1 for \(\dfrac{49}{4}\) (o.e.); \(2^{\text{nd}}\) M1 for subtracting \([E(T)]^2\) — \(-\dfrac{9}{4}\) is M0 but \(-\dfrac{9}{16}\) could be M1 |
| \(= \dfrac{49}{4} - \dfrac{81}{16} = 7\dfrac{3}{16}\) or \(\dfrac{115}{16}\) (o.e.) | A1 | AWRT 7.19; full marks can still be scored in (d) and (e) if \(a\) is incorrect |
| (4 marks) | Total 14 marks |
# Question 7:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(R=3 \cap B=0) = \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16}$ | M1, A1 | For $\dfrac{1}{4} \times \dfrac{1}{4}$ |
**(2 marks)**
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| All 0s correct in table | B1 | |
| All 1, 2, 3s correct in table | B1 | |
| All 4, 6, 9s correct in table | B1 | |
**(3 marks)**
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = \dfrac{7}{16}$ | B1 | For $\dfrac{7}{16}$ |
| $b = c = d = \dfrac{1}{16}$ | B1, B1 | $2^{\text{nd}}$ B1 for only one error in $b,c,d$; $3^{\text{rd}}$ B1 all of $b,c,d = \dfrac{1}{16}$ |
**(3 marks)**
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(T) = \left(1\times\dfrac{1}{16}\right)+\left(2\times\dfrac{1}{8}\right)+\left(3\times\dfrac{3}{8}\right)+\left(4\times\dfrac{1}{16}\right)+\ldots$ | M1 | For attempting $\sum tP(T=t)$, 3 or more terms correct or correct ft. NB dividing by a number other than 1 scores M0 |
| $= 2\dfrac{1}{4}$ or exact equivalent e.g. $2.25$, $\dfrac{9}{4}$ | A1 | |
**(2 marks)**
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(T) = \left(1^2\times\dfrac{1}{16}\right)+\left(2^2\times\dfrac{1}{8}\right)+\left(3^2\times\dfrac{3}{8}\right)+\left(4^2\times\dfrac{1}{16}\right)+\ldots-\left(\dfrac{9}{4}\right)^2$ | M1A1, M1 | $1^{\text{st}}$ M1 for attempt at $E(T^2)$, 3+ terms correct; A1 for $\dfrac{49}{4}$ (o.e.); $2^{\text{nd}}$ M1 for subtracting $[E(T)]^2$ — $-\dfrac{9}{4}$ is M0 but $-\dfrac{9}{16}$ could be M1 |
| $= \dfrac{49}{4} - \dfrac{81}{16} = 7\dfrac{3}{16}$ or $\dfrac{115}{16}$ (o.e.) | A1 | AWRT 7.19; full marks can still be scored in (d) and (e) if $a$ is incorrect |
**(4 marks) | Total 14 marks**7. Tetrahedral dice have four faces. Two fair tetrahedral dice, one red and one blue, have faces numbered $0,1,2$, and 3 respectively. The dice are rolled and the numbers face down on the two dice are recorded. The random variable $R$ is the score on the red die and the random variable $B$ is the score on the blue die.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( R = 3$ and $B = 0 )$.
The random variable $T$ is $R$ multiplied by $B$.
\item Complete the diagram below to represent the sample space that shows all the possible values of $T$.\\
\includegraphics[max width=\textwidth, alt={}, center]{af84d17b-5308-4b1e-99b5-40c5df5bf01e-13_732_771_834_621}
\section*{Sample space diagram of $T$}
\item The table below represents the probability distribution of the random variable $T$.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
$t$ & 0 & 1 & 2 & 3 & 4 & 6 & 9 \\
\hline
$\mathrm { P } ( T = t )$ & $a$ & $b$ & $1 / 8$ & $1 / 8$ & $c$ & $1 / 8$ & $d$ \\
\hline
\end{tabular}
\end{center}
Find the values of $a , b , c$ and $d$.
Find the values of
\item $\mathrm { E } ( T )$,
\item $\operatorname { Var } ( T )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q7 [14]}}