| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring basic probability calculations and E(X) computation. Part (i) is a simple chart drawing, part (ii) involves counting favorable outcomes from 36 equally likely dice pairs (routine probability argument), and part (iii) is direct application of E(X) = Σr·P(X=r) with simple fractions. All techniques are standard textbook exercises with no problem-solving insight required. |
| Spec | 2.04a Discrete probability distributions |
| \(r\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 6 }\) | \(\frac { 5 } { 18 }\) | \(\frac { 2 } { 9 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 9 }\) | \(\frac { 1 } { 18 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Bar chart with labelled linear scales on both axes, correct heights | G1, G1 | Accept \(r\) or \(x\) for horizontal label; \(p\) or probability for vertical. Bars must not be wider than gaps. Condone vertical scale 1,2,3,4,5 and Probability \(\times \frac{1}{18}\) as label. BOD for height of \(r=0\) on vertical axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(X=1\), possible scores are \((1,2),(2,3),(3,4),(4,5),(5,6)\) and \((2,1),(3,2),(4,3),(5,4),(6,5)\). All equally likely so probability \(= \frac{10}{36} = \frac{5}{18}\) | M1, A1 | M1 for clear correct sample space with ten 1's identified. No additional values such as 0,1 and 1,0. Do not allow 'just 10 ways you can have a difference of 1' so 10/36 or equivalent. SC1 for possible scores \((1,2),(2,3),(3,4),(4,5),(5,6)\) so probability \(= 2 \times 5 \times \frac{1}{36}\) with no explanation for \(2\times\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(X=0\), possible scores are \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\) so probability \(= \frac{6}{36} = \frac{1}{6}\) | B1 | B1 for clear correct sample space with six 0's identified. Allow both dice must be the same so probability \(= \frac{6}{36} = \frac{1}{6}\). Allow \(1 \times \frac{1}{6} = \frac{1}{6}\) BOD |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 0\times\frac{1}{6}+1\times\frac{5}{18}+2\times\frac{2}{9}+3\times\frac{1}{6}+4\times\frac{1}{9}+5\times\frac{1}{18} = 1\frac{17}{18} = 1.94\) | M1 for \(\Sigma rp\) (at least 3 terms correct), A1 CAO | Or \(\frac{35}{18}\). Division by 6 or other spurious factor gets MAX M1A0 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Bar chart with labelled linear scales on both axes, correct heights | G1, G1 | Accept $r$ or $x$ for horizontal label; $p$ or probability for vertical. Bars must not be wider than gaps. Condone vertical scale 1,2,3,4,5 and Probability $\times \frac{1}{18}$ as label. BOD for height of $r=0$ on vertical axis |
## Part (ii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $X=1$, possible scores are $(1,2),(2,3),(3,4),(4,5),(5,6)$ and $(2,1),(3,2),(4,3),(5,4),(6,5)$. All equally likely so probability $= \frac{10}{36} = \frac{5}{18}$ | M1, A1 | M1 for clear correct sample space with ten 1's identified. No additional values such as 0,1 and 1,0. Do not allow 'just 10 ways you can have a difference of 1' so 10/36 or equivalent. SC1 for possible scores $(1,2),(2,3),(3,4),(4,5),(5,6)$ so probability $= 2 \times 5 \times \frac{1}{36}$ with no explanation for $2\times$ |
## Part (ii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $X=0$, possible scores are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ so probability $= \frac{6}{36} = \frac{1}{6}$ | B1 | B1 for clear correct sample space with six 0's identified. Allow both dice must be the same so probability $= \frac{6}{36} = \frac{1}{6}$. Allow $1 \times \frac{1}{6} = \frac{1}{6}$ BOD |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 0\times\frac{1}{6}+1\times\frac{5}{18}+2\times\frac{2}{9}+3\times\frac{1}{6}+4\times\frac{1}{9}+5\times\frac{1}{18} = 1\frac{17}{18} = 1.94$ | M1 for $\Sigma rp$ (at least 3 terms correct), A1 CAO | Or $\frac{35}{18}$. Division by 6 or other spurious factor gets MAX M1A0 |
---2 Two fair six-sided dice are thrown. The random variable $X$ denotes the difference between the scores on the two dice. The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 6 }$ & $\frac { 5 } { 18 }$ & $\frac { 2 } { 9 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 9 }$ & $\frac { 1 } { 18 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Draw a vertical line chart to illustrate the probability distribution.
\item Use a probability argument to show that\\
(A) $\mathrm { P } ( X = 1 ) = \frac { 5 } { 18 }$,\\
(B) $\mathrm { P } ( X = 0 ) = \frac { 1 } { 6 }$.
\item Find the mean value of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q2 [7]}}