| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential selection without replacement |
| Difficulty | Easy -1.2 This is a straightforward tree diagram question requiring basic probability calculations with replacement-adjusted denominators (12→11) and simple conditional probability. The setup is clearly structured, requiring only routine application of standard S1 techniques with no problem-solving insight needed. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct probabilities on first set of branches: \(\frac{7}{12}\), \(\frac{3}{12}\), \(\frac{2}{12}\) | B1 | For remaining probs on first set of branches and at least one on 2nd set |
| Fully correct tree diagram with all correct probabilities (second counter branches showing \(\frac{6}{11}\), \(\frac{3}{11}\), \(\frac{2}{11}\) etc.) | B1 | For a fully correct tree diagram with all correct probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(Y) = \frac{7}{12}\times\frac{2}{11} + \frac{3}{12}\times\frac{2}{11} + \frac{2}{12} = \frac{42}{132}\) or \(\frac{7}{22}\) or \(P(\text{Yellow and two counters}) = \frac{7}{12}\times\frac{2}{11} + \frac{3}{12}\times\frac{2}{11} = \frac{20}{132}\) or \(\frac{5}{33}\) | M1 | For a correct ft expression for \(P(Y)\) or \(P(\text{Yellow and two counters})\) ft their tree diagram. eg \(1 - \frac{7}{12}\times\frac{6+3}{11} - \frac{3}{12}\times\frac{7+2}{11}\) |
| \(\frac{P([Y\cap R]\cup[Y\cap B])}{P(Y)} = \frac{\frac{20}{132}}{\frac{42}{132}}\) | M1 | For a correct ratio formula (symbols or words) and at least one correct ft prob or fully correct ft ratio. Do not follow through probabilities \(> 1\) or \(< 0\) |
| \(= \frac{20}{42}\) or \(\frac{10}{21}\) | A1 | For \(\frac{10}{21}\) or exact equivalent. Allow \(0.\dot{4}7619\dot{0}\). If exact correct fraction not given and awrt 0.476... given it would get M1M1A0 if from correct working |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct probabilities on first set of branches: $\frac{7}{12}$, $\frac{3}{12}$, $\frac{2}{12}$ | B1 | For remaining probs on first set of branches and at least one on 2nd set |
| Fully correct tree diagram with all correct probabilities (second counter branches showing $\frac{6}{11}$, $\frac{3}{11}$, $\frac{2}{11}$ etc.) | B1 | For a fully correct tree diagram with all correct probabilities |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(Y) = \frac{7}{12}\times\frac{2}{11} + \frac{3}{12}\times\frac{2}{11} + \frac{2}{12} = \frac{42}{132}$ or $\frac{7}{22}$ **or** $P(\text{Yellow and two counters}) = \frac{7}{12}\times\frac{2}{11} + \frac{3}{12}\times\frac{2}{11} = \frac{20}{132}$ or $\frac{5}{33}$ | M1 | For a correct ft expression for $P(Y)$ or $P(\text{Yellow and two counters})$ ft their tree diagram. eg $1 - \frac{7}{12}\times\frac{6+3}{11} - \frac{3}{12}\times\frac{7+2}{11}$ |
| $\frac{P([Y\cap R]\cup[Y\cap B])}{P(Y)} = \frac{\frac{20}{132}}{\frac{42}{132}}$ | M1 | For a correct ratio formula (symbols or words) and at least one correct ft prob or fully correct ft ratio. Do not follow through probabilities $> 1$ or $< 0$ |
| $= \frac{20}{42}$ or $\frac{10}{21}$ | A1 | For $\frac{10}{21}$ or exact equivalent. Allow $0.\dot{4}7619\dot{0}$. If exact correct fraction not given and awrt 0.476... given it would get M1M1A0 if from correct working |
---\begin{enumerate}
\item There are 7 red counters, 3 blue counters and 2 yellow counters in a bag. Gina selects a counter at random from the bag and keeps it. If the counter is yellow she does not select any more counters. If the counter is not yellow she randomly selects a second counter from the bag.\\
(a) Complete the tree diagram.
\end{enumerate}
First Counter\\
Second Counter\\
\includegraphics[max width=\textwidth, alt={}, center]{a439724e-b570-434d-bf75-de2b50915042-02_1147_1081_603_397}
Given that Gina has selected a yellow counter,\\
(b) find the probability that she has 2 counters.\\
\hfill \mbox{\textit{Edexcel S1 2021 Q1 [5]}}