Moderate -0.8 This is a straightforward S1 question requiring basic probability distribution skills: drawing a simple chart, counting favorable outcomes from a 6×6 sample space (standard dice problem), and calculating E(X) using the given distribution. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average A-level maths.
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 \(X\).
\(r\)
0
1
2
3
4
5
P(X = r)
\(\frac{1}{6}\)
\(\frac{5}{18}\)
\(\frac{2}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{9}\)
\(\frac{1}{18}\)
Draw a vertical line chart to illustrate the probability distribution. [2]
Answer/Working: G1 labelled linear scales on both axes, G1 heights
Answer
Marks
Marks:
2
Guidance: Accept \(r\) or \(x\) for horizontal label and \(p\) or better for vertical including probability distribution. Visual check only. Allow G1G0 for points rather than lines. Bars must not be wider than gaps for second G1. Condone vertical scale 1, 2, 3, 4, 5 and Probability (\(\times\)) 1/18 as label. BOD for height of \(r = 0\) on vertical axis
(ii)
Answer/Working: (A) 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 are equally likely) so probability = \(\frac{10}{36} = \frac{5}{18}\)
Answer
Marks
Marks: M1, A1
2
Guidance: Also M1 for a clear correct sample space seen with the ten 1's identified by means of circles or ticks oe soi. Must be convincing. 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 are (1,2), (2,3), (3,4), (4,5), (5,6) so probability = \(2 \times 5 \times 1/36\) with no explanation for \(2 \times\)
Answer/Working: (B) If \(X = 0\), possible scores are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) so probability = \(\frac{6}{36} = \frac{1}{6}\)
Answer
Marks
Marks: B1
1
Guidance: Also B1 for a clear correct sample space seen with the six 0's identified by means of circles or ticks oe soi. Must be convincing. No additional values. Allow both dice must be the same so probability = 6/36 = 1/6. Allow \(1 \times 1/6= 1/6\) BOD
Marks: M1 for \(\Sigma rp\) (at least 3 terms correct), A1 CAO
2
Guidance: Or 35/18. Division by 6 or other spurious factor gets MAX M1A0
## (i)
**Answer/Working:** G1 labelled linear scales on both axes, G1 heights
**Marks:** | 2
**Guidance:** Accept $r$ or $x$ for horizontal label and $p$ or better for vertical including probability distribution. Visual check only. Allow G1G0 for points rather than lines. Bars must not be wider than gaps for second G1. Condone vertical scale 1, 2, 3, 4, 5 and Probability ($\times$) 1/18 as label. BOD for height of $r = 0$ on vertical axis
## (ii)
**Answer/Working:** (A) 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 are equally likely) so probability = $\frac{10}{36} = \frac{5}{18}$
**Marks:** M1, A1 | 2
**Guidance:** Also M1 for a clear correct sample space seen with the ten 1's identified by means of circles or ticks oe soi. Must be convincing. 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 are (1,2), (2,3), (3,4), (4,5), (5,6) so probability = $2 \times 5 \times 1/36$ with no explanation for $2 \times$
**Answer/Working:** (B) If $X = 0$, possible scores are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) so probability = $\frac{6}{36} = \frac{1}{6}$
**Marks:** B1 | 1
**Guidance:** Also B1 for a clear correct sample space seen with the six 0's identified by means of circles or ticks oe soi. Must be convincing. No additional values. Allow both dice must be the same so probability = 6/36 = 1/6. Allow $1 \times 1/6= 1/6$ BOD
## (iii)
**Answer/Working:** Mean value of $X = 0 \times \frac{1}{6} + 1 \times \frac{5}{18} + 2 \times \frac{2}{9} + 3 \times \frac{1}{6} + 4 \times \frac{1}{9} + 5 \times \frac{1}{18} = \frac{17}{18} = 1.94$
**Marks:** M1 for $\Sigma rp$ (at least 3 terms correct), A1 CAO | 2
**Guidance:** Or 35/18. Division by 6 or other spurious factor gets MAX M1A0
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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{tabular}{|c|c|c|c|c|c|c|}
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
P(X = r) & $\frac{1}{6}$ & $\frac{5}{18}$ & $\frac{2}{9}$ & $\frac{1}{6}$ & $\frac{1}{9}$ & $\frac{1}{18}$ \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Draw a vertical line chart to illustrate the probability distribution. [2]
\item Use a probability argument to show that
\begin{enumerate}[label=(\Alph*)]
\item P(X = 1) = $\frac{5}{18}$. [2]
\item P(X = 0) = $\frac{1}{6}$. [1]
\end{enumerate}
\item Find the mean value of $X$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2011 Q4 [7]}}