| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes (12 equally likely cases), construction of a probability table, and calculation of variance using standard formulas. While it requires careful organization and multiple steps, it involves only routine S1 techniques with no conceptual challenges or novel insights—slightly easier than average due to its mechanical nature. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(x\) values: \(0, 2, 4, 6, 8, 10, 12\) and \(P(X=x)\): \(\frac{3}{12}, \frac{2}{12}, \frac{2}{12}, \frac{2}{12}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}\) | B1 | Table with correct \(x\) values and at least 2 correct probabilities. Condone any additional \(x\) values if probability stated as 0 |
| B1 | Four more probabilities correctly linked to the correct \(x\) value, need not be in table, accept unsimplified | |
| B1 | 7 correct probabilities linked with correct outcomes. Decimals correct to at least 3 SF. SC B1 for 7 or more probabilities summing to 1 placed in a probability distribution table |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 0 + 2\times\frac{2}{12} + 4\times\frac{2}{12} + 6\times\frac{2}{12} + 8\times\frac{1}{12} + 10\times\frac{1}{12} + 12\times\frac{1}{12}\) \(= \frac{54}{12} = 4.5\) | M1 | Accept unsimplified expression. Accept \(\frac{1}{3}+\frac{2}{3}+1+\frac{2}{3}+\frac{5}{6}+1\) OE for M mark FT their table with 7 or 8 probabilities summing to \(0.999 \leqslant \text{total} \leqslant 1\), \((0 < p < 1)\) |
| \(\text{Var}(X) = 0 + 2^2\times\frac{2}{12} + 4^2\times\frac{2}{12} + 6^2\times\frac{2}{12} + 8^2\times\frac{1}{12} + 10^2\times\frac{1}{12} + 12^2\times\frac{1}{12} - (4.5)^2\) \(= 35 - 4.5^2\) | M1 | Appropriate variance formula using their \((E(X))^2\) value. FT their table with 6 or more probabilities \((0 < p < 1)\) |
| \(= 14.75,\ 14\frac{3}{4}\) | A1 | CAO. Accept \(\frac{59}{4}\). If either/both M marks not awarded, SC B1 for correct answer WWW |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $x$ values: $0, 2, 4, 6, 8, 10, 12$ and $P(X=x)$: $\frac{3}{12}, \frac{2}{12}, \frac{2}{12}, \frac{2}{12}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}$ | B1 | Table with correct $x$ values and at least 2 correct probabilities. Condone any additional $x$ values if probability stated as 0 |
| | B1 | Four more probabilities correctly linked to the correct $x$ value, need not be in table, accept unsimplified |
| | B1 | 7 correct probabilities linked with correct outcomes. Decimals correct to at least 3 SF. SC B1 for 7 or more probabilities summing to 1 placed in a probability distribution table |
**Total: 3 marks**
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 0 + 2\times\frac{2}{12} + 4\times\frac{2}{12} + 6\times\frac{2}{12} + 8\times\frac{1}{12} + 10\times\frac{1}{12} + 12\times\frac{1}{12}$ $= \frac{54}{12} = 4.5$ | M1 | Accept unsimplified expression. Accept $\frac{1}{3}+\frac{2}{3}+1+\frac{2}{3}+\frac{5}{6}+1$ OE for M mark FT their table with 7 or 8 probabilities summing to $0.999 \leqslant \text{total} \leqslant 1$, $(0 < p < 1)$ |
| $\text{Var}(X) = 0 + 2^2\times\frac{2}{12} + 4^2\times\frac{2}{12} + 6^2\times\frac{2}{12} + 8^2\times\frac{1}{12} + 10^2\times\frac{1}{12} + 12^2\times\frac{1}{12} - (4.5)^2$ $= 35 - 4.5^2$ | M1 | Appropriate variance formula using their $(E(X))^2$ value. FT their table with 6 or more probabilities $(0 < p < 1)$ |
| $= 14.75,\ 14\frac{3}{4}$ | A1 | CAO. Accept $\frac{59}{4}$. If either/both M marks not awarded, SC B1 for correct answer WWW |
**Total: 3 marks**3 A fair coin and an ordinary fair six-sided dice are thrown at the same time.The random variable $X$ is defined as follows.\\
-If the coin shows a tail,$X$ is twice the score on the dice.\\
-If the coin shows a head,$X$ is the score on the dice if the score is even and $X$ is 0 otherwise.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$ .
\item Find $\operatorname { Var } ( X )$ .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q3 [6]}}