| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This question requires constructing a probability distribution by systematically considering cases (same vs different values) and applying basic probability rules for independent events. Part (a) involves organized enumeration and part (b) uses straightforward conditional probability. While requiring careful bookkeeping across multiple cases, it's a standard S1 exercise with no novel insight needed—slightly easier than average due to the small sample space and routine application of independence. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(y\): \(1, 2, 3, 4\) and prob: \(\frac{7}{16}, \frac{5}{16}, \frac{3}{16}, \frac{1}{16}\) | B1 | Probability distribution table with correct scores with at least one probability; allow extra score values if probability of zero stated |
| B1 | One probability (linked with correct score) correct | |
| B1 | 2 more probabilities (linked with correct scores) correct | |
| B1 FT | \(4\)th probability correct, FT sum of 3 or 4 terms \(= 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2\ | \text{even}) = \dfrac{\frac{5}{16}}{\frac{6}{16}}\) | M1 |
| \(\frac{5}{6}\) or \(0.833\) | A1 |
## Question 4:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $y$: $1, 2, 3, 4$ and prob: $\frac{7}{16}, \frac{5}{16}, \frac{3}{16}, \frac{1}{16}$ | B1 | Probability distribution table with correct scores with at least one probability; allow extra score values if probability of zero stated |
| | B1 | One probability (linked with correct score) correct |
| | B1 | 2 more probabilities (linked with correct scores) correct |
| | B1 FT | $4$th probability correct, FT sum of 3 or 4 terms $= 1$ |
**Total: 4 marks**
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2\|\text{even}) = \dfrac{\frac{5}{16}}{\frac{6}{16}}$ | M1 | $\dfrac{\text{their } P(2)}{\text{their } P(2) + \text{their } P(4)}$ seen or correct outcome space |
| $\frac{5}{6}$ or $0.833$ | A1 | |
**Total: 2 marks**
---4 The random variable $X$ takes each of the values $1,2,3,4$ with probability $\frac { 1 } { 4 }$. Two independent values of $X$ are chosen at random. If the two values of $X$ are the same, the random variable $Y$ takes that value. Otherwise, the value of $Y$ is the larger value of $X$ minus the smaller value of $X$.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $Y$.
\item Find the probability that $Y = 2$ given that $Y$ is even.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q4 [6]}}