| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two independent categorical choices |
| Difficulty | Easy -1.3 This is a straightforward tree diagram question requiring basic probability rules: completing branches (probabilities sum to 1), using the law of total probability to find p, and applying conditional probability formulas. All steps are routine S1 techniques with no problem-solving insight needed, making it easier than average. |
| 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 |
|---|---|---|
| \(1-p\), \(\frac{7}{8}\) and \(\frac{9}{10}\) in correct place on tree diagram | B1 | For fully correct tree diagram with all 3 correct labels; allow if \(1-p\) is seen and crossed out/replaced with a numerical probability |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{8}p + \frac{1}{10}(1-p) = 0.11\) | M1 A1ft | For \(\frac{1}{8}p\) or \(\frac{1}{10}(1-p)\) seen in equation for \(p\); A1ft for fully correct equation in \(p\) or correct ft equation based on their tree diagram |
| \(p = \frac{2}{5}\) | A1 | correct answer scores 3 out of 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2}{5}\times\frac{1}{8} = \frac{1}{20}\) | M1 A1ft | For \(p\times\frac{1}{8}\) ft their \(p\), provided \(p\) is a probability; correct answer scores 2 out of 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y12 | R) = \dfrac{\frac{2}{5}\times\frac{7}{8}}{1-0.11}\) or \(P(Y12 | R) = \dfrac{\frac{2}{5}\times\frac{7}{8}}{\frac{2}{5}\times\frac{7}{8}+\frac{3}{5}\times\frac{9}{10}}\) |
| \(= \frac{35}{89}\) | A1 | For \(\frac{35}{89}\) (allow awrt 0.393) |
## Question 3:
### Part (a):
$1-p$, $\frac{7}{8}$ and $\frac{9}{10}$ in correct place on tree diagram | B1 | For fully correct tree diagram with all 3 correct labels; allow if $1-p$ is seen and crossed out/replaced with a numerical probability
### Part (b):
$\frac{1}{8}p + \frac{1}{10}(1-p) = 0.11$ | M1 A1ft | For $\frac{1}{8}p$ or $\frac{1}{10}(1-p)$ seen in equation for $p$; A1ft for fully correct equation in $p$ or correct ft equation based on their tree diagram
$p = \frac{2}{5}$ | A1 | correct answer scores 3 out of 3
### Part (c):
$\frac{2}{5}\times\frac{1}{8} = \frac{1}{20}$ | M1 A1ft | For $p\times\frac{1}{8}$ ft their $p$, provided $p$ is a probability; correct answer scores 2 out of 2
### Part (d):
$P(Y12|R) = \dfrac{\frac{2}{5}\times\frac{7}{8}}{1-0.11}$ or $P(Y12|R) = \dfrac{\frac{2}{5}\times\frac{7}{8}}{\frac{2}{5}\times\frac{7}{8}+\frac{3}{5}\times\frac{9}{10}}$ | M1 | For correct ratio of probabilities; can ft their $p$, provided $p$ is a probability
$= \frac{35}{89}$ | A1 | For $\frac{35}{89}$ (allow awrt 0.393)
---\begin{enumerate}
\item In a sixth form college each student in Year 12 and Year 13 is either left-handed (L) or right-handed (R).
\end{enumerate}
The partially completed tree diagram, where $p$ is a probability, gives information about these students.\\
\includegraphics[max width=\textwidth, alt={}, center]{86446ce3-496a-4f02-9566-9b207bac9efa-10_960_981_477_543}\\
(a) Complete the tree diagram, in terms of $p$ where necessary.
The probability that a student is left-handed is 0.11\\
(b) Find the value of $p$\\
(c) Find the probability that a student selected at random is in Year 12 and left-handed.
Given that a student is right-handed,\\
(d) find the probability that the student is in Year 12
\hfill \mbox{\textit{Edexcel S1 2024 Q3 [8]}}