| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sampling without replacement tree diagram |
| Difficulty | Easy -1.2 This is a standard textbook exercise on sampling without replacement with straightforward probability calculations. Drawing the tree diagram and computing conditional probabilities using basic fraction arithmetic requires only routine application of well-practiced techniques with no problem-solving insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram drawn with correct structure | M1 | |
| \(\frac{9}{12}, \frac{3}{12}\) on first branches; \(\frac{8}{11}, \frac{3}{11}\) on second branches from Blue | A1 | |
| \(\frac{9}{11}, \frac{2}{11}\) on second branches from Red; complete and labelled | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Second ball is red}) = \frac{9}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{2}{11} = \frac{1}{4}\) | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Both red} \mid \text{Second ball is red}) = \dfrac{\frac{3}{12} \times \frac{2}{11}}{\frac{1}{4}} = \frac{2}{11}\) | M1A | exact or awrt 0.182 |
# Question 4:
## Part (a)
| Tree diagram drawn with correct structure | M1 | |
| $\frac{9}{12}, \frac{3}{12}$ on first branches; $\frac{8}{11}, \frac{3}{11}$ on second branches from Blue | A1 | |
| $\frac{9}{11}, \frac{2}{11}$ on second branches from Red; complete and labelled | A1 | (3) |
## Part (b)
| $P(\text{Second ball is red}) = \frac{9}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{2}{11} = \frac{1}{4}$ | M1A1 | (2) |
## Part (c)
| $P(\text{Both red} \mid \text{Second ball is red}) = \dfrac{\frac{3}{12} \times \frac{2}{11}}{\frac{1}{4}} = \frac{2}{11}$ | M1A | exact or awrt 0.182 | (2) |
---4. A bag contains 9 blue balls and 3 red balls. A ball is selected at random from the bag and its colour is recorded. The ball is not replaced. A second ball is selected at random and its colour is recorded.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to represent the information.
Find the probability that\\
(a) the second ball selected is red,
\item both balls selected are red, given that the second ball selected is red.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2006 Q4 [7]}}