| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Sampling without replacement from bags/boxes |
| Difficulty | Standard +0.3 This is a standard conditional probability question with tree diagrams and straightforward algebraic manipulation. Part (a) requires basic probability fractions, part (b) is routine algebra showing a given result, and part (c) applies Bayes' theorem in a familiar context. The question is slightly easier than average due to its structured guidance and standard techniques. |
| 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 |
| Probabilities: \(\frac{x+1}{x+10}\), \(\frac{9}{x+10}\), \(\frac{x}{x+10}\), \(\frac{10}{x+10}\) | B1 | One probability correct in correct position |
| B1 | Another probability correct in correct position | |
| B1 | Other two probabilities correct in correct positions | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4}{10} \times their\ \frac{10}{x+10}\) | M1 | Method consistent with *their* tree diagram |
| \(\frac{4}{x+10}\) | A1 | AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4}{x+10} = \frac{1}{6}\), so \(x+10=24\), \(x=14\) | B1 | Find value of \(x\). Can be implied by correct probabilities in calculation |
| \(P(A\text{Red}\mid B\text{Red}) = P(A\text{Red} \cap B\text{Red}) \div P(B\text{Red})\) | ||
| \(\dfrac{\frac{6}{10} \times their\ \frac{x+1}{x+10}}{\frac{6}{10} \times their\ \frac{x+1}{x+10} + \frac{4}{10} \times their\ \frac{x}{x+10}} = \dfrac{\frac{6}{10} \times \frac{15}{24}}{\frac{6}{10} \times \frac{15}{24} + \frac{4}{10} \times \frac{14}{24}} = \dfrac{\frac{3}{8}}{\frac{73}{120}}\) | B1 FT | \(\frac{6}{10} \times their\ \frac{x+1}{x+10}\) as numerator or denominator of fraction |
| M1 | \(\frac{6}{10} \times their\ \frac{x+1}{x+10} + \frac{4}{10} \times their\ \frac{x}{x+10}\) seen anywhere | |
| A1 FT | Seen as denominator of fraction | |
| \(\frac{45}{73}\), \(0.616[4\ldots]\) | A1 | If B0 M0: SC B1 for \(\dfrac{\frac{3}{8}}{\frac{73}{120}}\) or \(\dfrac{0.375}{0.6083}\); SC B1 \(\frac{45}{73}\) or \(0.616\) |
| 5 |
## Question 7:
### Part 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probabilities: $\frac{x+1}{x+10}$, $\frac{9}{x+10}$, $\frac{x}{x+10}$, $\frac{10}{x+10}$ | **B1** | One probability correct in correct position |
| | **B1** | Another probability correct in correct position |
| | **B1** | Other two probabilities correct in correct positions |
| | **3** | |
---
### Part 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{10} \times their\ \frac{10}{x+10}$ | **M1** | Method consistent with *their* tree diagram |
| $\frac{4}{x+10}$ | **A1** | AG |
| | **2** | |
---
### Part 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{x+10} = \frac{1}{6}$, so $x+10=24$, $x=14$ | **B1** | Find value of $x$. Can be implied by correct probabilities in calculation |
| $P(A\text{Red}\mid B\text{Red}) = P(A\text{Red} \cap B\text{Red}) \div P(B\text{Red})$ | | |
| $\dfrac{\frac{6}{10} \times their\ \frac{x+1}{x+10}}{\frac{6}{10} \times their\ \frac{x+1}{x+10} + \frac{4}{10} \times their\ \frac{x}{x+10}} = \dfrac{\frac{6}{10} \times \frac{15}{24}}{\frac{6}{10} \times \frac{15}{24} + \frac{4}{10} \times \frac{14}{24}} = \dfrac{\frac{3}{8}}{\frac{73}{120}}$ | **B1 FT** | $\frac{6}{10} \times their\ \frac{x+1}{x+10}$ as numerator or denominator of fraction |
| | **M1** | $\frac{6}{10} \times their\ \frac{x+1}{x+10} + \frac{4}{10} \times their\ \frac{x}{x+10}$ seen anywhere |
| | **A1 FT** | Seen as denominator of fraction |
| $\frac{45}{73}$, $0.616[4\ldots]$ | **A1** | If B0 M0: **SC B1** for $\dfrac{\frac{3}{8}}{\frac{73}{120}}$ or $\dfrac{0.375}{0.6083}$; **SC B1** $\frac{45}{73}$ or $0.616$ |
| | **5** | |7 Box $A$ contains 6 red balls and 4 blue balls. Box $B$ contains $x$ red balls and 9 blue balls. A ball is chosen at random from box $A$ and placed in box $B$. A ball is then chosen at random from box $B$.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram below, giving the remaining four probabilities in terms of $x$.\\
\includegraphics[max width=\textwidth, alt={}, center]{217c5a58-2966-4b86-b3b6-9d1676d2979c-12_688_759_484_731}
\item Show that the probability that both balls chosen are blue is $\frac { 4 } { x + 10 }$.\\
It is given that the probability that both balls chosen are blue is $\frac { 1 } { 6 }$.
\item Find the probability, correct to 3 significant figures, that the ball chosen from box $A$ is red given that the ball chosen from box $B$ is red.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q7 [10]}}