| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two-stage sampling with replacement |
| Difficulty | Moderate -0.8 This is a straightforward two-stage probability question with replacement involving tree diagrams. Part (a) requires basic probability calculations (7/8, 1/8, then conditional probabilities like 10/15, 9/15, etc.). Part (b) is standard multiplication and addition of probabilities. Part (c) involves conditional probability using Bayes' theorem, which is a routine S1 technique. All steps are mechanical applications of standard formulas with no conceptual challenges or novel insights required. |
| 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 |
| Box A: \(P(\text{Red}) = \frac{7}{8}\), \(P(\text{Blue}) = \frac{1}{8}\) | B1 | Both correct probs, box A |
| Box B: \(P(\text{Red}\mid\text{Red from A}) = \frac{10}{15}\), \(P(\text{Blue}\mid\text{Red from A}) = \frac{5}{15}\) | B1 | 2 probs correct for box B |
| Box B: \(P(\text{Red}\mid\text{Blue from A}) = \frac{9}{15}\), \(P(\text{Blue}\mid\text{Blue from A}) = \frac{6}{15}\) | B1 | All correct probs for box B |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{9}{15}\) | M1 | Two 2-factor terms added, correct or FT *their* 6(a) |
| \(= \frac{44}{120} \left[\frac{11}{30} \text{ or } 0.367\right]\) | A1 | OE |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{A blue} \mid \text{B blue}) = \dfrac{P(\text{A blue} \cap \text{B blue})}{P(\text{B blue})}\) | M1 | *their* \(\dfrac{1}{8} \times \dfrac{6}{15}\) seen as numerator or denominator of fraction |
| \(= \dfrac{\dfrac{1}{8} \times \dfrac{6}{15}}{\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}} = \dfrac{\dfrac{1}{20}}{\dfrac{41}{120}}\) | M1 | *their* \(\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}\) seen |
| M1 | *their* \(\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}\) seen as denominator | |
| \(= \dfrac{6}{41}\) or \(0.146\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Box A: $P(\text{Red}) = \frac{7}{8}$, $P(\text{Blue}) = \frac{1}{8}$ | B1 | Both correct probs, box A |
| Box B: $P(\text{Red}\mid\text{Red from A}) = \frac{10}{15}$, $P(\text{Blue}\mid\text{Red from A}) = \frac{5}{15}$ | B1 | 2 probs correct for box B |
| Box B: $P(\text{Red}\mid\text{Blue from A}) = \frac{9}{15}$, $P(\text{Blue}\mid\text{Blue from A}) = \frac{6}{15}$ | B1 | All correct probs for box B |
| **Total: 3** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{9}{15}$ | M1 | Two 2-factor terms added, correct or FT *their* **6(a)** |
| $= \frac{44}{120} \left[\frac{11}{30} \text{ or } 0.367\right]$ | A1 | OE |
| **Total: 2** | | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{A blue} \mid \text{B blue}) = \dfrac{P(\text{A blue} \cap \text{B blue})}{P(\text{B blue})}$ | M1 | *their* $\dfrac{1}{8} \times \dfrac{6}{15}$ seen as numerator or denominator of fraction |
| $= \dfrac{\dfrac{1}{8} \times \dfrac{6}{15}}{\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}} = \dfrac{\dfrac{1}{20}}{\dfrac{41}{120}}$ | M1 | *their* $\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}$ seen |
| | M1 | *their* $\dfrac{7}{8} \times \dfrac{5}{15} + \dfrac{1}{8} \times \dfrac{6}{15}$ seen as denominator |
| $= \dfrac{6}{41}$ or $0.146$ | A1 | |
---6 Box $A$ contains 7 red balls and 1 blue ball. Box $B$ contains 9 red balls and 5 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 $B$. The tree diagram below shows the possibilities for the colours of the balls chosen.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram to show the probabilities.
Box $A$\\
\includegraphics[max width=\textwidth, alt={}, center]{f7c0e35d-1889-4e5b-b094-f467052a66cf-08_624_428_667_621}
\section*{Box $B$}
Red
Blue
Red
Blue
\item Find the probability that the two balls chosen are not the same colour.
\item Find the probability that the ball chosen from box $A$ is blue given that the ball chosen from box $B$ is blue.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q6 [9]}}