| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| 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 and basic algebraic manipulation. Part (a) requires constructing a standard tree diagram with conditional probabilities, while part (b) involves setting up and solving a simple equation using the total probability formula. The algebra is uncomplicated (linear equation in x), and the question follows a very standard template for S1 level work. |
| 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 |
| Tree diagram: Box A branches with probabilities \(\frac{6}{9}\) and \(\frac{3}{9}\) | B1 | Correct structure and probabilities for Box A branches |
| Box B branch with \(\frac{5}{5+x}\) for G, \(\frac{x}{5+x}\) for Y (upper) | B1 | Completely correct structure and one correct probability for a Box B branch including label for G or Y |
| Second correct probability on a Box B branch | B1 | Completely correct structure and second correct probability on a Box B branch including label for G or Y |
| Remaining two probabilities correct: \(\frac{4}{5+x}\) for G, \(\frac{1+x}{5+x}\) for Y (lower) | B1 | Completely correct structure and remaining two probabilities correct on Box B branches. SC B1 if correct shape but only four correct algebraic probs for GG, GY, YG and YY |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{same colour}) = \frac{6}{9}\times\frac{5}{5+x}+\frac{3}{9}\times\frac{1+x}{5+x}\) | M1 | \(P(GG)+P(YY) = \left(\frac{6}{9}\text{ or }\frac{3}{9}\right)\times their\!\left(\frac{5}{5+x}\right)+\left(\frac{3}{9}\text{ or }\frac{6}{9}\right)\times their\!\left(\frac{1+x}{5+x}\right)\) |
| \(\frac{6}{9}\times\frac{5}{5+x}+\frac{3}{9}\times\frac{1+x}{5+x} = \frac{8}{15}\) and arrange as linear equation | M1 | \(15(x+11)=24(x+5)\) OE. Accept sum of their products equated to \(\frac{8}{15}\) and rearranged to linear equation |
| Solve: \(x = 5\) | A1 |
# Question 3:
## Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram: Box A branches with probabilities $\frac{6}{9}$ and $\frac{3}{9}$ | B1 | Correct structure and probabilities for Box A branches |
| Box B branch with $\frac{5}{5+x}$ for G, $\frac{x}{5+x}$ for Y (upper) | B1 | Completely correct structure and one correct probability for a Box B branch including label for G or Y |
| Second correct probability on a Box B branch | B1 | Completely correct structure and second correct probability on a Box B branch including label for G or Y |
| Remaining two probabilities correct: $\frac{4}{5+x}$ for G, $\frac{1+x}{5+x}$ for Y (lower) | B1 | Completely correct structure and remaining two probabilities correct on Box B branches. **SC B1** if correct shape but only four correct algebraic probs for GG, GY, YG and YY |
## Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{same colour}) = \frac{6}{9}\times\frac{5}{5+x}+\frac{3}{9}\times\frac{1+x}{5+x}$ | M1 | $P(GG)+P(YY) = \left(\frac{6}{9}\text{ or }\frac{3}{9}\right)\times their\!\left(\frac{5}{5+x}\right)+\left(\frac{3}{9}\text{ or }\frac{6}{9}\right)\times their\!\left(\frac{1+x}{5+x}\right)$ |
| $\frac{6}{9}\times\frac{5}{5+x}+\frac{3}{9}\times\frac{1+x}{5+x} = \frac{8}{15}$ and arrange as linear equation | M1 | $15(x+11)=24(x+5)$ OE. Accept sum of their products equated to $\frac{8}{15}$ and rearranged to linear equation |
| Solve: $x = 5$ | A1 | |
---3 Box $A$ contains 6 green balls and 3 yellow balls.\\
Box $B$ contains 4 green balls and $x$ yellow 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 Draw a tree diagram to represent this information, showing the probability on each of the branches.\\[0pt]
[4]\\
\includegraphics[max width=\textwidth, alt={}, center]{9b21cc0f-b043-4251-8aa9-cb1e5c2fb5d0-06_2727_38_132_2010}
\begin{center}
\end{center}
The probability that both the balls chosen are the same colour is $\frac { 8 } { 15 }$.
\item Find the value of $x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q3 [7]}}