| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Moderate -0.3 This is a straightforward S1 probability question requiring systematic enumeration of outcomes and basic conditional probability. Part (a) and (b) involve listing ways to achieve scores (e.g., T=2 from rolling 0+2, 1+1, or 2+0), then calculating probabilities using the given distribution. Part (c) applies conditional probability with a clear condition. While multi-part and requiring careful organization, it uses only basic probability rules without novel insight—slightly easier than average A-level. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.04a Discrete probability distributions |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(P(T = 2) = 3 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{12}\) oe | M1 A1 |
| (b) | \(P(T = 3) = \left[P(0, 3) + P(1, 2) + P(2, 1)\right] + P(3)\) | M1M1 |
| \(= \left(\frac{1}{6} \times \frac{1}{2}\right) + \left(\frac{1}{6} \times \frac{1}{6}\right) + \left(\frac{1}{6} \times \frac{1}{2}\right) + \frac{1}{2}\) | ||
| \(= \frac{23}{36}\) oe | A1 | A1 allow exact equivalent |
| (c) | \(P(T = 3 \mid \text{rolled twice}) = \frac{P((T=3) \cap \text{die rolled twice})}{P(\text{die rolled twice})}\) | M1 |
| \(= \frac{5}{36}\) | M1 | 2nd M1 for correct numerical ratio of probabilities (allow ft of (their (b) − \(\frac{1}{2}\)) as numerator) |
| \(= \frac{1}{2}\) | ||
| \(= \frac{5}{18}\) oe | A1 | A1 allow exact equivalent |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $P(T = 2) = 3 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{12}$ oe | M1 A1 | M1 for correct expression. A1 allow exact equivalent |
| (b) | $P(T = 3) = \left[P(0, 3) + P(1, 2) + P(2, 1)\right] + P(3)$ | M1M1 | 1st M1 for $\frac{1}{2} +$ at least one correct product. 2nd M1 for fully correct expression |
| | $= \left(\frac{1}{6} \times \frac{1}{2}\right) + \left(\frac{1}{6} \times \frac{1}{6}\right) + \left(\frac{1}{6} \times \frac{1}{2}\right) + \frac{1}{2}$ | | |
| | $= \frac{23}{36}$ oe | A1 | A1 allow exact equivalent |
| (c) | $P(T = 3 \mid \text{rolled twice}) = \frac{P((T=3) \cap \text{die rolled twice})}{P(\text{die rolled twice})}$ | M1 | 1st M1 for correct conditional probability ratio (this mark may be implied by 2nd M1) but going on to assume independence [using numerator $P(T=3) \times P(\text{rolled twice})$] is M0M0A0. |
| | $= \frac{5}{36}$ | M1 | 2nd M1 for correct numerical ratio of probabilities (allow ft of (their (b) − $\frac{1}{2}$) as numerator) |
| | $= \frac{1}{2}$ | | |
| | $= \frac{5}{18}$ oe | A1 | A1 allow exact equivalent |
**Total: 8 marks**
**Correct answer only in (a), (b) or (c) scores full marks for that part. Methods leading to answers > 1 score 0 marks.**
---5. A biased tetrahedral die has faces numbered $0,1,2$ and 3 . The die is rolled and the number face down on the die, $X$, is recorded. The probability distribution of $X$ is
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 2 }$ \\
\hline
\end{tabular}
\end{center}
If $X = 3$ then the final score is 3\\
If $X \neq 3$ then the die is rolled again and the final score is the sum of the two numbers.
The random variable $T$ is the final score.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( T = 2 )$
\item Find $\mathrm { P } ( T = 3 )$
\item Given that the die is rolled twice, find the probability that the final score is 3
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q5 [8]}}