| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Transfer between containers |
| Difficulty | Moderate -0.3 This is a standard S1 tree diagram question with conditional probability. While it requires careful tracking of changing bag compositions across multiple stages and uses conditional probability in part (d), the methodology is routine for this topic. The calculations are straightforward once the tree diagram is set up correctly, making it slightly easier than average for A-level. |
| 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 | Mark | Guidance |
| \(\frac{5}{8}\) and \(\frac{3}{8}\) on 2nd branches | B1 | Allow 0.625 & 0.375 or 62.5% & 37.5% |
| \(\frac{8}{13}\) and \(\frac{5}{13}\) on 3rd branches | B1 | Allow awrt 0.615 & awrt 0.385 |
| \(\frac{7}{13}\) and \(\frac{6}{13}\) on 3rd branches | B1 | Allow awrt 0.538 & awrt 0.462 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5}{9}\times\frac{4}{8} + \frac{4}{9}\times\frac{5}{8} = \frac{5}{9}\) | M1 A1 | For \(\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{5}{8}\) ft tree; allow 0.556 or 55.6% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13} + \frac{4}{9}\times\frac{3}{8}\times\frac{7}{13} = \frac{61}{234}\) | M1 A1 | Allow awrt 0.261 or 26.1% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13}}{\frac{61}{234}} = \frac{\frac{20}{117}}{\frac{61}{234}} = \frac{40}{61}\) | M1 A1ft A1 | For \(\frac{\text{a probability}}{\text{part (c)}}\) where numerator < denominator; allow awrt 0.656 or 65.6% |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{8}$ and $\frac{3}{8}$ on 2nd branches | B1 | Allow 0.625 & 0.375 or 62.5% & 37.5% |
| $\frac{8}{13}$ and $\frac{5}{13}$ on 3rd branches | B1 | Allow awrt 0.615 & awrt 0.385 |
| $\frac{7}{13}$ and $\frac{6}{13}$ on 3rd branches | B1 | Allow awrt 0.538 & awrt 0.462 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{9}\times\frac{4}{8} + \frac{4}{9}\times\frac{5}{8} = \frac{5}{9}$ | M1 A1 | For $\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{5}{8}$ ft tree; allow 0.556 or 55.6% |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13} + \frac{4}{9}\times\frac{3}{8}\times\frac{7}{13} = \frac{61}{234}$ | M1 A1 | Allow awrt 0.261 or 26.1% |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13}}{\frac{61}{234}} = \frac{\frac{20}{117}}{\frac{61}{234}} = \frac{40}{61}$ | M1 A1ft A1 | For $\frac{\text{a probability}}{\text{part (c)}}$ where numerator < denominator; allow awrt 0.656 or 65.6% |
---\begin{enumerate}
\item Two bags, $\boldsymbol { X }$ and $\boldsymbol { Y }$, each contain green marbles (G) and blue marbles (B) only.
\end{enumerate}
\begin{itemize}
\item Bag $\boldsymbol { X }$ contains 5 green marbles and 4 blue marbles
\item Bag $\boldsymbol { Y }$ contains 6 green marbles and 5 blue marbles
\end{itemize}
A marble is selected at random from bag $\boldsymbol { X }$ and placed in bag $\boldsymbol { Y }$\\
A second marble is selected at random from bag $\boldsymbol { X }$ and placed in bag $\boldsymbol { Y }$\\
A third marble is then selected, this time from bag $\boldsymbol { Y }$\\
(a) Use this information to complete the tree diagram shown on page 7\\
(b) Find the probability that the 2 marbles selected from bag $\boldsymbol { X }$ are of different colours.\\
(c) Find the probability that all 3 marbles selected are the same colour.
Given that all three marbles selected are the same colour,\\
(d) find the probability that they are all green.
2nd Marble (from bag $\boldsymbol { X }$ )
\section*{3rd Marble (from bag Y)}
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{1st Marble (from bag $\boldsymbol { X }$ )}
\includegraphics[alt={},max width=\textwidth]{c316fa29-dedc-4890-bd82-31eb0bb819f9-07_1694_1312_484_310}
\end{center}
\end{figure}
\hfill \mbox{\textit{Edexcel S1 2023 Q2 [10]}}