| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | March |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability with replacement vs without replacement |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question with clear structure. Part (a) requires a simple two-step probability calculation (3/12 × 2/11). Part (b) involves Bayes' theorem or a tree diagram, which is standard S1 material. The replacement rule is clearly stated, making this slightly easier than average but still requiring careful probability reasoning. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\frac{3}{12} \times \frac{2}{11} =\right] \frac{1}{22}\) | B1 | Accept \(\frac{6}{132}\), \(0.0454\dot{5}\)... to at least three significant figures. |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(B1 \mid R2) = \dfrac{\frac{9}{12} \times \frac{3}{12}}{\frac{9}{12} \times \frac{3}{12} + \frac{3}{12} \times \frac{2}{11}}\) | M1 | \(\frac{9}{12} \times \frac{3}{12} = \frac{27}{144}\), \(\frac{3}{16}\), \(0.1875\) seen as numerator or denominator of a fraction. |
| M1 | Their \(\frac{9}{12} \times \frac{3}{12} +\) their \(\left(\frac{3}{12} \times \frac{2}{11}\right)\) or their \(\frac{27}{144} +\) their \(\frac{1}{22}\) seen as denominator of a fraction. FT from part (a). | |
| \(= \dfrac{\frac{3}{16}}{\frac{3}{16} + \frac{1}{22}} = \frac{33}{41}\), \(0.805\) | A1 | Accept \(\frac{4752}{5904}\) oe, \(0.804878\)... rounded to at least three significant figures. If A0, SC B1 for correct final answer www. |
| 3 |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{3}{12} \times \frac{2}{11} =\right] \frac{1}{22}$ | **B1** | Accept $\frac{6}{132}$, $0.0454\dot{5}$... to at least three significant figures. |
| | **1** | |
---
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(B1 \mid R2) = \dfrac{\frac{9}{12} \times \frac{3}{12}}{\frac{9}{12} \times \frac{3}{12} + \frac{3}{12} \times \frac{2}{11}}$ | **M1** | $\frac{9}{12} \times \frac{3}{12} = \frac{27}{144}$, $\frac{3}{16}$, $0.1875$ seen as numerator or denominator of a fraction. |
| | **M1** | Their $\frac{9}{12} \times \frac{3}{12} +$ their $\left(\frac{3}{12} \times \frac{2}{11}\right)$ or their $\frac{27}{144} +$ their $\frac{1}{22}$ seen as denominator of a fraction. FT from part **(a)**. |
| $= \dfrac{\frac{3}{16}}{\frac{3}{16} + \frac{1}{22}} = \frac{33}{41}$, $0.805$ | **A1** | Accept $\frac{4752}{5904}$ oe, $0.804878$... rounded to at least three significant figures. If A0, **SC B1** for correct final answer www. |
| | **3** | |1 A bag contains 9 blue marbles and 3 red marbles. One marble is chosen at random from the bag. If this marble is blue, it is replaced back into the bag. If this marble is red, it is not returned to the bag. A second marble is now chosen at random from the bag.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that both the marbles chosen are red.
\item Find the probability that the first marble chosen is blue given that the second marble chosen is red.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q1 [4]}}