| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability with replacement vs without replacement |
| Difficulty | Standard +0.3 This is a standard conditional probability question with clear structure: part (a) requires a tree diagram with replacement vs without replacement branches, and part (b) applies Bayes' theorem. The calculations are straightforward with simple fractions, and the setup explicitly guides students through the two scenarios. This is slightly easier than average because the problem structure is transparent and the arithmetic is manageable. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(HRR) = \frac{1}{4}\times\frac{4}{6}\times\frac{4}{6} = \frac{16}{144}, \frac{4}{36}\) \(P(TRR) = \frac{3}{4}\times\frac{3}{7}\times\frac{2}{6} = \frac{18}{168}, \frac{3}{28}\) \(P(HBB) = \frac{1}{4}\times\frac{2}{6}\times\frac{2}{6} = \frac{4}{144}, \frac{1}{36}\) \(P(TBB) = \frac{3}{4}\times\frac{4}{7}\times\frac{3}{6} = \frac{36}{168}, \frac{6}{28}, \frac{3}{14}\) | B1 | 2 clearly identified unsimplified probabilities from \(P(HRR)\), \(P(TRR)\), \(P(HBB)\) and \(P(TBB)\) correct |
| \(\frac{4}{36} + \frac{3}{28} + \frac{1}{36} + \frac{6}{28}\) | M1 | Sum of 4 correct scenarios, may be identified by the unsimplified probability calculations |
| \(= \frac{29}{63}\) or \(0.460\) | A1 | \(0.460317\ldots\) to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(T | BB) = \frac{P(T \cap BB)}{P(BB)} = \frac{\frac{3}{4}\times\frac{4}{7}\times\frac{3}{6}}{\frac{1}{36}+\frac{6}{28}}\) | M1 |
| M1 | Their \(\frac{1}{36}\) + their \(\frac{6}{28}\) FT from 4(a) or correct, \(0.24206\ldots\), seen as denominator of a fraction; accept unsimplified | |
| \(\frac{3}{14} \div \frac{61}{252}\) or \(\frac{\frac{3}{14}}{\frac{61}{252}}\) | ||
| \(\frac{54}{61}\) or \(0.885\) | A1 | Accept \(0.8852589\ldots\) rounded to at least 3SF. If one or both Ms not awarded, SC B1 for correct final answer WWW |
# Question 4:
## Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(HRR) = \frac{1}{4}\times\frac{4}{6}\times\frac{4}{6} = \frac{16}{144}, \frac{4}{36}$ $P(TRR) = \frac{3}{4}\times\frac{3}{7}\times\frac{2}{6} = \frac{18}{168}, \frac{3}{28}$ $P(HBB) = \frac{1}{4}\times\frac{2}{6}\times\frac{2}{6} = \frac{4}{144}, \frac{1}{36}$ $P(TBB) = \frac{3}{4}\times\frac{4}{7}\times\frac{3}{6} = \frac{36}{168}, \frac{6}{28}, \frac{3}{14}$ | B1 | 2 clearly identified unsimplified probabilities from $P(HRR)$, $P(TRR)$, $P(HBB)$ and $P(TBB)$ correct |
| $\frac{4}{36} + \frac{3}{28} + \frac{1}{36} + \frac{6}{28}$ | M1 | Sum of 4 correct scenarios, may be identified by the unsimplified probability calculations |
| $= \frac{29}{63}$ or $0.460$ | A1 | $0.460317\ldots$ to at least 3SF |
## Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(T|BB) = \frac{P(T \cap BB)}{P(BB)} = \frac{\frac{3}{4}\times\frac{4}{7}\times\frac{3}{6}}{\frac{1}{36}+\frac{6}{28}}$ | M1 | $\frac{3}{4}\times\frac{4}{7}\times\frac{3}{6}, \frac{36}{168}$, oe, $\frac{3}{14}$, $0.2142857\ldots$ seen as numerator of a fraction; accept unsimplified; FT their $P(TBB)$ from **4(a)** |
| | M1 | Their $\frac{1}{36}$ + their $\frac{6}{28}$ FT from **4(a)** or correct, $0.24206\ldots$, seen as denominator of a fraction; accept unsimplified |
| $\frac{3}{14} \div \frac{61}{252}$ or $\frac{\frac{3}{14}}{\frac{61}{252}}$ | | |
| $\frac{54}{61}$ or $0.885$ | A1 | Accept $0.8852589\ldots$ rounded to at least 3SF. If one or both Ms not awarded, **SC B1** for correct final answer WWW |
---4 Rahul has two bags, $X$ and $Y$. Bag $X$ contains 4 red marbles and 2 blue marbles. Bag $Y$ contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is $\frac { 1 } { 4 }$.
Rahul throws the coin.
\begin{itemize}
\item If he obtains a head, he chooses at random a marble from bag $X$. He notes the colour and replaces the marble in bag $X$. He then chooses at random a second marble from bag $X$.
\item If he obtains a tail, he chooses at random a marble from bag $Y$. He notes the colour and discards the marble. He then chooses at random a second marble from bag $Y$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the two marbles that Rahul chooses are the same colour.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-06_2717_33_109_2014}\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-07_2725_35_99_20}
\item Find the probability that the two marbles that Rahul chooses are both from bag $Y$ given that both marbles are blue.\\
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q4 [6]}}