| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Moderate -0.8 This is a straightforward application of the independence definition P(A∩B) = P(A)×P(B) using basic probability counting with dice. Students need to enumerate outcomes systematically but the conceptual demand is low—it's a standard textbook exercise testing recall of the independence criterion rather than requiring problem-solving insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Complete outcome space shown (6×6 grid with sums from 2 to 12), or listing A and B outcomes, or listing \(A \cap B\) outcomes | M1 | Complete outcome space or listing A and B outcomes or listing \(A \cap B\) outcomes |
| \(P(A \cap B) = \dfrac{5}{36}\) | A1 | With evidence |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(A) \times P(B) = \frac{1}{3} \times \frac{10}{36}\) | M1 | Their \(\frac{1}{3} \times\) their \(\frac{10}{36}\) seen |
| \(\frac{5}{54} \neq \frac{5}{36}\) so not independent | A1 | \(\frac{5}{54}, \frac{5}{36}\), \(P(A) \times P(B)\) and \(P(A \cap B)\) seen in workings and correct conclusion stated. Condone \(\frac{5}{36}\) being stated in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(B\ | A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{5}{36}}{\frac{1}{3}}\) | M1 |
| \(\frac{5}{12} \neq \frac{5}{18}\) so not independent | A1 | \(P(A\ |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Complete outcome space shown (6×6 grid with sums from 2 to 12), or listing A and B outcomes, or listing $A \cap B$ outcomes | M1 | Complete outcome space or listing A and B outcomes or listing $A \cap B$ outcomes |
| $P(A \cap B) = \dfrac{5}{36}$ | A1 | With evidence |
| | **Total: 2** | |
## Question 1:
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A) \times P(B) = \frac{1}{3} \times \frac{10}{36}$ | M1 | Their $\frac{1}{3} \times$ their $\frac{10}{36}$ seen |
| $\frac{5}{54} \neq \frac{5}{36}$ so not independent | A1 | $\frac{5}{54}, \frac{5}{36}$, $P(A) \times P(B)$ and $P(A \cap B)$ seen in workings and correct conclusion stated. Condone $\frac{5}{36}$ being stated in **(a)** |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(B\|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{5}{36}}{\frac{1}{3}}$ | M1 | OE, $\frac{\text{their } 1(a)}{\text{their } P(A)}$ seen |
| $\frac{5}{12} \neq \frac{5}{18}$ so not independent | A1 | $P(A\|B)$, $P(B)$, $\frac{5}{12}, \frac{5}{18}$ seen in workings and correct conclusion stated. Condone $\frac{5}{18} = \frac{10}{36}$ being identified in **(a)** |
**Total: 2 marks**
---1 Two ordinary fair dice, one red and the other blue, are thrown.\\
Event $A$ is 'the score on the red die is divisible by 3 '.\\
Event $B$ is 'the sum of the two scores is at least 9 '.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( A \cap B )$.
\item Hence determine whether or not the events $A$ and $B$ are independent.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q1 [4]}}