| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.8 This is a straightforward application of tree diagrams and conditional probability with clearly stated probabilities. Parts (a)-(c) are routine S1 exercises requiring basic probability rules (multiplication along branches, addition of paths, Bayes' theorem). Part (d) adds minimal complexity by asking for the complement case. No conceptual insight or problem-solving is needed beyond standard textbook methods. |
| 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 with correct number of branches | M1 | |
| Probabilities 0.2, 0.3, 0.5 correct | A1 | |
| All probabilities correct | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{used sauna}) = (0.2 \times 0.35) + (0.3 \times 0.2) + (0.5 \times 0.45) = 0.355\) | M1 A1 A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{swim} \mid \text{sauna used}) = \frac{P(\text{swim \& sauna})}{P(\text{sauna})} = \frac{0.2 \times 0.35}{0.355} = 0.19718\) | M1 A1 A1 (3) | Accept awrt 0.197 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{swim} \mid \text{sauna not used}) = \frac{P(\text{sauna not used} \mid \text{swim})\,P(\text{swim})}{P(\text{sauna not used})}\) | M1 | |
| \(P(\text{sauna not used} \mid \text{swim}) = 1 - 0.35 = 0.65\) | B1 | |
| \(P(\text{sauna not used}) = 1 - 0.355 = 0.645\) | M1 A1 f.t. | |
| \(\therefore P(\text{swim} \mid \text{sauna not used}) = \frac{0.65 \times 0.2}{0.645} = 0.20155\) | M1 A1 (6) | Accept awrt 0.202 (15) |
## Question 5:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram with correct number of branches | M1 | |
| Probabilities 0.2, 0.3, 0.5 correct | A1 | |
| All probabilities correct | A1 (3) | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{used sauna}) = (0.2 \times 0.35) + (0.3 \times 0.2) + (0.5 \times 0.45) = 0.355$ | M1 A1 A1 (3) | |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{swim} \mid \text{sauna used}) = \frac{P(\text{swim \& sauna})}{P(\text{sauna})} = \frac{0.2 \times 0.35}{0.355} = 0.19718$ | M1 A1 A1 (3) | Accept awrt 0.197 |
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{swim} \mid \text{sauna not used}) = \frac{P(\text{sauna not used} \mid \text{swim})\,P(\text{swim})}{P(\text{sauna not used})}$ | M1 | |
| $P(\text{sauna not used} \mid \text{swim}) = 1 - 0.35 = 0.65$ | B1 | |
| $P(\text{sauna not used}) = 1 - 0.355 = 0.645$ | M1 A1 f.t. | |
| $\therefore P(\text{swim} \mid \text{sauna not used}) = \frac{0.65 \times 0.2}{0.645} = 0.20155$ | M1 A1 (6) | Accept awrt 0.202 **(15)** |
---5. A keep-fit enthusiast swims, runs or cycles each day with probabilities $0.2,0.3$ and 0.5 respectively. If he swims he then spends time in the sauna with probability 0.35 . The probabilities that he spends time in the sauna after running or cycling are 0.2 and 0.45 respectively.
\begin{enumerate}[label=(\alph*)]
\item Represent this information on a tree diagram.
\item Find the probability that on any particular day he uses the sauna.
\item Given that he uses the sauna one day, find the probability that he had been swimming.
\item Given that he did not use the sauna one day, find the probability that he had been swimming.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [15]}}