| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring systematic enumeration of outcomes from a non-standard die. Part (a) involves listing 36 equally likely outcomes and counting frequencies—routine but tedious. Parts (b) and (c) are standard variance calculation and conditional probability. The question requires careful bookkeeping rather than conceptual insight, making it easier than average A-level material. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(x\) values: \(2, 3, 4, 5, 6\) and \(P(X=x)\): \(\frac{1}{36}, \frac{4}{36}, \frac{10}{36}, \frac{12}{36}, \frac{9}{36}\) — correct \(X\) value with at least one correct associated probability | B1 | Table with correct \(X\) values and at least one correct probability associated with the correct \(X\) value. Values need not be in order, lines may not be drawn, may be vertical, \(X\) and \(P(X)\) may be omitted. Condone any additional \(X\) values if probability stated as 0. |
| Simplified forms: \(\frac{1}{36}, \frac{1}{9}, \frac{5}{18}, \frac{1}{3}, \frac{1}{4}\) — three other probabilities associated with correct \(x\) values | B1 | Three other probabilities associated with correct \(x\) values, need not be in table, accept unsimplified. |
| Decimal equivalents (3sf): \(0.0278, 0.111, 0.278, 0.333, 0.25\) — five correct probabilities linked with correct outcomes | B1 | Five correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for five probabilities summing to 1 placed in a probability distribution table with the correct \(x\) values. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{1\times2+4\times3+10\times4+12\times5+9\times6}{36} = \frac{2}{36}+\frac{12}{36}+\frac{40}{36}+\frac{60}{36}+\frac{54}{36} = \frac{14}{3}\) or 4.67 | M1 | Accept unsimplified expression or sum of fractions. FT their table with five probabilities summing to \(0.999 \leqslant total \leqslant 1\) |
| \(\text{Var}(X) = \frac{1\times2^2+4\times3^2+10\times4^2+12\times5^2+9\times6^2}{36} - \left(\frac{14}{3}\right)^2\) | M1 | Appropriate variance formula using their \((E(X))^2\) value. FT their table with 4 or more probabilities. Note: \(\left(\frac{824}{36}\) or \(\frac{206}{9}\) or \(22.89\right)-\left(\frac{196}{9}\) or \(21.78\) or \(\left(\frac{14}{3}\right)^2\right)\) implies M1 |
| \(\left[= \frac{824}{36} - \frac{196}{9} = 22.89 - 21.78\right] = \frac{10}{9}\) | A1 | \(1\frac{1}{9}\), 1.11[1], 1.1̇ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \text{ even} \mid X>3) = \dfrac{\frac{10}{36}+\frac{9}{36}}{\frac{31}{36}}\) | M1 | \(\frac{their\ P(4)+their\ P(6)}{their\ P(4)+P(5)+P(6)}\), all probabilities \((0 < p < 1)\). If sample space seen in any part of question, then M1 \(\frac{their\ 19}{their\ 31}\) |
| \(= \frac{19}{31}\) | A1 | 0.613 |
## Question 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $x$ values: $2, 3, 4, 5, 6$ and $P(X=x)$: $\frac{1}{36}, \frac{4}{36}, \frac{10}{36}, \frac{12}{36}, \frac{9}{36}$ — correct $X$ value with at least one correct associated probability | **B1** | Table with correct $X$ values and at least one correct probability associated with the correct $X$ value. Values need not be in order, lines may not be drawn, may be vertical, $X$ and $P(X)$ may be omitted. Condone any additional $X$ values if probability stated as 0. |
| Simplified forms: $\frac{1}{36}, \frac{1}{9}, \frac{5}{18}, \frac{1}{3}, \frac{1}{4}$ — three other probabilities associated with correct $x$ values | **B1** | Three other probabilities associated with correct $x$ values, need not be in table, accept unsimplified. |
| Decimal equivalents (3sf): $0.0278, 0.111, 0.278, 0.333, 0.25$ — five correct probabilities linked with correct outcomes | **B1** | Five correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. **SC B1** for five probabilities summing to 1 placed in a probability distribution table with the correct $x$ values. |
| | **[3]** | |
# Question 1:
## Part 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{1\times2+4\times3+10\times4+12\times5+9\times6}{36} = \frac{2}{36}+\frac{12}{36}+\frac{40}{36}+\frac{60}{36}+\frac{54}{36} = \frac{14}{3}$ or 4.67 | M1 | Accept unsimplified expression or sum of fractions. FT their table with five probabilities summing to $0.999 \leqslant total \leqslant 1$ |
| $\text{Var}(X) = \frac{1\times2^2+4\times3^2+10\times4^2+12\times5^2+9\times6^2}{36} - \left(\frac{14}{3}\right)^2$ | M1 | Appropriate variance formula using their $(E(X))^2$ value. FT their table with 4 or more probabilities. Note: $\left(\frac{824}{36}$ or $\frac{206}{9}$ or $22.89\right)-\left(\frac{196}{9}$ or $21.78$ or $\left(\frac{14}{3}\right)^2\right)$ implies M1 |
| $\left[= \frac{824}{36} - \frac{196}{9} = 22.89 - 21.78\right] = \frac{10}{9}$ | A1 | $1\frac{1}{9}$, 1.11[1], 1.1̇ |
## Part 1(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \text{ even} \mid X>3) = \dfrac{\frac{10}{36}+\frac{9}{36}}{\frac{31}{36}}$ | M1 | $\frac{their\ P(4)+their\ P(6)}{their\ P(4)+P(5)+P(6)}$, all probabilities $(0 < p < 1)$. If sample space seen in any part of question, then M1 $\frac{their\ 19}{their\ 31}$ |
| $= \frac{19}{31}$ | A1 | 0.613 |
---1 The numbers on the faces of a fair six-sided dice are $1,2,2,3,3,3$. The random variable $X$ is the total score when the dice is rolled twice.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$.
\item Find the value of $\operatorname { Var } ( X )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b21cc0f-b043-4251-8aa9-cb1e5c2fb5d0-02_2714_34_143_2012}
\item Find the probability that $X$ is even given that $X > 3$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q1 [8]}}