| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sampling without replacement |
| Difficulty | Moderate -0.8 This is a straightforward sampling without replacement problem requiring basic counting of outcomes (5 coins choose 2 = 10 equally likely pairs), constructing a simple probability distribution table, and applying the standard variance formula. All steps are routine S1 techniques with no conceptual challenges or novel insights required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Or \([\text{Probability} =]\ 0.2 \times 0.5 \times 2 = 0.2\) | B1 | AG. Must include [\\(7], 5, 2 and link the probabilities to the appropriate value. \)\dfrac{[{}^1C_1] \times {}^2C_1}{{}^5C_2} = 0.2\(. \)\frac{1}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4}\(, not \)\frac{1}{5} \times \frac{2}{4} + \frac{1}{5} \times \frac{2}{4}$ unless 5 and 2 and 2 and 5 seen in solution. If all possibilities identified (e.g. outcome table), must be clearly labelled and terms fulfilling the condition identified. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | 2 | 3 |
| \(P(X=x)\) | 0.1 | 0.4 |
| B1 | Table with correct \(x\) values and at least one further non-zero probability correct. Condone extra \(x\) values if probability stated as 0. | |
| B1 | Two more correct non-zero probabilities linked with correct outcomes. Accept probabilities not in table if clearly identified. | |
| B1 | All five probabilities correct. Accept probabilities not in table if clearly identified. SC B1 for four further non-zero probabilities adding to 0.8 if B1 max scored. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 0.1\times2 + 0.4\times3 + 0.1\times4 + 0.2\times6 + 0.2\times7\) \(= 0.2+1.2+0.4+1.2+1.4 = 4.4\) | M1 | Accept unsimplified expression. FT their table with at least 5 probabilities, \(0 < p < 1\), summing to 1. FT acceptable at bold partially evaluated stage. |
| \(Var(X) = 0.1\times2^2+0.4\times3^2+0.1\times4^2+0.2\times6^2+0.2\times7^2-4.4^2\) \(= 0.1\times4+0.4\times9+0.1\times16+0.2\times36+0.2\times49-4.4^2\) | M1 | Appropriate variance formula using their \((E(X))^2\). FT table with at least 4 probabilities, \(0 < p < 1\). Note: if table correct, \(22.6-(4.4^2\) or \(19.36)\) implies this M1. |
| \(= 3.24\) | A1 | CAO. \(\frac{81}{25}\), \(3\frac{6}{25}\) scores A0. Only dependent on previous M1. If M0 M0 scored, SC B1 for 3.24 WWW. |
## Question 5(a):
$[\$7 =]\ [\$]5 + [\$]2$
$[\text{Probability} =]\ \dfrac{1}{5} \times \dfrac{2}{4} \times 2 = \dfrac{1}{5} = 0.2$
Or $[\text{Probability} =]\ 0.2 \times 0.5 \times 2 = 0.2$ | B1 | AG. Must include [\$7], 5, 2 and link the probabilities to the appropriate value. $\dfrac{[{}^1C_1] \times {}^2C_1}{{}^5C_2} = 0.2$. $\frac{1}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4}$, not $\frac{1}{5} \times \frac{2}{4} + \frac{1}{5} \times \frac{2}{4}$ unless 5 and 2 and 2 and 5 seen in solution. If all possibilities identified (e.g. outcome table), must be clearly labelled and terms fulfilling the condition identified.
**Total: 1 mark**
---
## Question 5(b):
| $x$ | 2 | 3 | 4 | 6 | 7 |
|---|---|---|---|---|---|
| $P(X=x)$ | 0.1 | 0.4 | 0.1 | 0.2 | 0.2 |
| B1 | Table with correct $x$ values and at least one further non-zero probability correct. Condone extra $x$ values if probability stated as 0.
| B1 | Two more correct non-zero probabilities linked with correct outcomes. Accept probabilities not in table if clearly identified.
| B1 | All five probabilities correct. Accept probabilities not in table if clearly identified. **SC B1** for four further non-zero probabilities adding to 0.8 if B1 max scored.
**Total: 3 marks**
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 0.1\times2 + 0.4\times3 + 0.1\times4 + 0.2\times6 + 0.2\times7$ $= 0.2+1.2+0.4+1.2+1.4 = 4.4$ | M1 | Accept unsimplified expression. FT their table with at least 5 probabilities, $0 < p < 1$, summing to 1. FT acceptable at bold partially evaluated stage. |
| $Var(X) = 0.1\times2^2+0.4\times3^2+0.1\times4^2+0.2\times6^2+0.2\times7^2-4.4^2$ $= 0.1\times4+0.4\times9+0.1\times16+0.2\times36+0.2\times49-4.4^2$ | M1 | Appropriate variance formula using their $(E(X))^2$. FT table with at least 4 probabilities, $0 < p < 1$. Note: if table correct, $22.6-(4.4^2$ or $19.36)$ implies this M1. |
| $= 3.24$ | A1 | CAO. $\frac{81}{25}$, $3\frac{6}{25}$ scores A0. Only dependent on previous M1. If M0 M0 scored, SC B1 for 3.24 WWW. |
---5 Jasmine has one $\$ 5$ coin, two $\$ 2$ coins and two $\$ 1$ coins. She selects two of these coins at random. The random variable $X$ is the total value, in dollars, of these two coins.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 7 ) = 0.2$.
\item Draw up the probability distribution table for $X$.
\item Find the value of $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q5 [7]}}