| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Standard +0.3 This is a straightforward S1 discrete probability distribution question with standard bookwork parts (a)-(d) requiring only direct application of formulas for expectation and variance. Parts (e)-(g) involve simple probability calculations with clear cases to enumerate. While multi-part, each step is routine and requires no novel insight—slightly easier than the typical A-level question due to its mechanical nature. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(s\) | 0 | 1 | 2 | 4 | 5 |
| \(\text{P}(S = s)\) | \(p\) | 0.25 | 0.25 | 0.20 | 0.20 |
| Answer | Marks |
|---|---|
| \(1 = p + (0.25 + 0.25 + 0.2 + 0.2)\), \(\Rightarrow p = \frac{1}{10}\) or 0.1 | M1, A1 |
| Answer | Marks |
|---|---|
| \(\text{E}(S) = \frac{1}{4} + 2 \times \frac{1}{4} + 4 \times \frac{1}{5} + 5 \times \frac{1}{5}\) (or equiv. in decimals) = 2.55 | M1, A1 |
| Answer | Marks |
|---|---|
| \(E\left(S^2\right) = \frac{1}{4} + \frac{2^2}{4} + \frac{4^2}{5} + \frac{5^2}{5}\) or 0.25 + 1 + 3.2 + 5 = 9.45 (*) | M1, A1cso |
| Answer | Marks |
|---|---|
| \(\text{Var}(S) = 9.45 - (\text{E}(S))^2\) = 2.9475 or \(\frac{1179}{400}\) (accept awrt 2.95) | M1, A1 |
| Answer | Marks |
|---|---|
| P(5 and 5) = \(\left(\frac{1}{5}\right)^2\) = \(\frac{1}{25}\) or 0.04 | M1, A1 |
| Answer | Marks |
|---|---|
| P(4, 4, 2) = \(\left(\frac{1}{5}\right)^2 \times \frac{1}{4} \times 3\) ( = 0.03 or \(\frac{3}{100}\)) | M1, M1 |
| P(4, 4, 4) = \(\left(\frac{1}{5}\right)^3\) ( = 0.008 or \(\frac{1}{125}\)) | B1 |
| P(Tom wins in 3 spins) = 0.038 | A1 |
| Answer | Marks |
|---|---|
| \(P\left(\bar{5} \cap \bar{5} \cap 5\right) + \text{P}(5 \cap \bar{5} \cap 5) = \frac{4}{5} \times \left(\frac{1}{5}\right)^2 \times 2 = \frac{0.064 \text{ or } \frac{8}{125}}\) | M1, M1, A1 |
## (a)
$1 = p + (0.25 + 0.25 + 0.2 + 0.2)$, $\Rightarrow p = \frac{1}{10}$ or 0.1 | M1, A1 |
## (b)
$\text{E}(S) = \frac{1}{4} + 2 \times \frac{1}{4} + 4 \times \frac{1}{5} + 5 \times \frac{1}{5}$ (or equiv. in decimals) = 2.55 | M1, A1 |
## (c)
$E\left(S^2\right) = \frac{1}{4} + \frac{2^2}{4} + \frac{4^2}{5} + \frac{5^2}{5}$ or 0.25 + 1 + 3.2 + 5 = 9.45 (*) | M1, A1cso |
## (d)
$\text{Var}(S) = 9.45 - (\text{E}(S))^2$ = 2.9475 or $\frac{1179}{400}$ (accept awrt 2.95) | M1, A1 |
## (e)
P(5 and 5) = $\left(\frac{1}{5}\right)^2$ = $\frac{1}{25}$ or 0.04 | M1, A1 |
## (f)
P(4, 4, 2) = $\left(\frac{1}{5}\right)^2 \times \frac{1}{4} \times 3$ ( = 0.03 or $\frac{3}{100}$) | M1, M1 |
P(4, 4, 4) = $\left(\frac{1}{5}\right)^3$ ( = 0.008 or $\frac{1}{125}$) | B1 |
P(Tom wins in 3 spins) = 0.038 | A1 |
## (g)
$P\left(\bar{5} \cap \bar{5} \cap 5\right) + \text{P}(5 \cap \bar{5} \cap 5) = \frac{4}{5} \times \left(\frac{1}{5}\right)^2 \times 2 = \frac{0.064 \text{ or } \frac{8}{125}}$ | M1, M1, A1 |
**Notes:**
- **(a)** M1 for clear attempt to use sum of probabilities = 1 (fractions or decimals). Ans only 2/2
- **(b)** M1 for at least 2 correct terms (≠ 0) of the expression, 2.55 with no working scores M1A1
- **Any division by k (usually 5) in (b) or (c) or (d) scores M0**
- **(c)** M1 for at least 3 correct, non-zero terms of the expression seen, allow decimals.
- A1cso for the full expression (with 9.45) seen. Must be cso but can ignore wrong p.
- **(d)** M1 for a correct expression (9.45 seen), can ft their E(S). May see $\sum (x - "2.55")^2 \times \text{P}(X = x)$
- A1 accept awrt 2.95. **Answer only can score M1 for correct ft and A1 for awrt 2.95**. **Answer only in (e) and (f) is full marks, in (g) is no marks**
- **(e)** M1 for $\left(\frac{1}{5}\right)^2$. Condone P(5) × P(5) = 0.25 × 0.25. [Beware 0.4 is A0]
- **(f)** 1st M1 for $\left(\frac{1}{5}\right)^2 \times \frac{1}{4}$ or 0.01 seen
- 2nd M1 for multiplying a p²q probability by 3(p, q ∈ (0,1)). B1 for (0.2)³ or better seen
- **(g)** 1st M1 for $\frac{4}{5} \times \left(\frac{1}{5}\right)^2$ or all cases considered and correct attempt at probabilities.
- 2nd M1 for multiplying a p²(1 − p) probability by 2. **Beware (0.4)³ = 0.064 is M0M0A0**A spinner is designed so that the score $S$ is given by the following probability distribution.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$s$ & 0 & 1 & 2 & 4 & 5 \\
\hline
$\text{P}(S = s)$ & $p$ & 0.25 & 0.25 & 0.20 & 0.20 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$. [2]
\item Find $\text{E}(S)$. [2]
\item Show that $\text{E}(S^2) = 9.45$ [2]
\item Find $\text{Var}(S)$. [2]
\end{enumerate}
Tom and Jess play a game with this spinner. The spinner is spun repeatedly and $S$ counters are awarded on the outcome of each spin. If $S$ is even then Tom receives the counters and if $S$ is odd then Jess receives them. The first player to collect 10 or more counters is the winner.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Find the probability that Jess wins after 2 spins. [2]
\item Find the probability that Tom wins after exactly 3 spins. [4]
\item Find the probability that Jess wins after exactly 3 spins. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2011 Q8 [17]}}