| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Markov chain weather transitions |
| Difficulty | Moderate -0.3 This is a straightforward Markov chain problem with clear transition probabilities. Part (i) is routine tree diagram construction, part (ii) involves standard probability calculations along branches, and part (iii) requires conditional probability using Bayes' theorem. While it has multiple parts and requires careful organization, all techniques are standard S1 material with no novel insights needed—slightly easier than average due to its mechanical nature. |
| 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 |
|---|---|---|
| G1 first set of branches, G1 indep second set of branches, G1 indep third set of branches, G1 labels | 4 marks | |
| (ii) (A) \(P(\text{all on time}) = 0.95^3 = 0.8574\) | M1 for \(0.95^3\), A1 CAO | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.019 + 0.0015 + 0.012 = 0.0325\) | M1 first term, M1 second term, M1 third term, A1 CAO | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.05 \times 0.4 \times 0.6 = 0.857375 + 0.0285 + 0.0285 + 0.012 = 0.926375\) | M1 any two terms, M1 third term, M1 fourth term, A1 CAO | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{P(\text{1000 on time and 1200 on time})}{P(\text{1200 on time})} = \frac{0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6}{0.926375} = \frac{0.885875}{0.926375} = 0.9563\) | M1 either term of numerator, M1 full numerator, M1 denominator, A1 CAO | 4 marks |
**(i)**
```
1200
/ \
0.95 On time
/
1100
/ \
0.95 On time 0.05 Late
/ / \
0.6 On time 0.4 Late
1000 / \
0.95 On time
/
0.05 Late
/ \
0.6 On time 0.4 Late
/ \
0.05 Late
/ \
0.6 On time
/ \
0.4 Late
```
| G1 first set of branches, G1 indep second set of branches, G1 indep third set of branches, G1 labels | 4 marks
**(ii) (A)** $P(\text{all on time}) = 0.95^3 = 0.8574$ | M1 for $0.95^3$, A1 CAO | 2 marks
**(B)** $P(\text{just one on time}) =$
$0.95 \times 0.05 \times 0.4 + 0.05 \times 0.6 \times 0.05 + 0.05 \times 0.4 \times 0.6$
$= 0.019 + 0.0015 + 0.012 = 0.0325$ | M1 first term, M1 second term, M1 third term, A1 CAO | 4 marks
**(C)** $P(\text{1200 is on time}) =$
$0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6 + 0.05 \times 0.6 \times 0.95 +$
$0.05 \times 0.4 \times 0.6 = 0.857375 + 0.0285 + 0.0285 + 0.012 = 0.926375$ | M1 any two terms, M1 third term, M1 fourth term, A1 CAO | 4 marks
**(iii)** $P(\text{1000 on time given 1200 on time}) =$
$\frac{P(\text{1000 on time and 1200 on time})}{P(\text{1200 on time})} = \frac{0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6}{0.926375} = \frac{0.885875}{0.926375} = 0.9563$ | M1 either term of numerator, M1 full numerator, M1 denominator, A1 CAO | 4 marks
**TOTAL: 18 marks**7 One train leaves a station each hour. The train is either on time or late. If the train is on time, the probability that the next train is on time is 0.95 . If the train is late, the probability that the next train is on time is 0.6 . On a particular day, the 0900 train is on time.
\begin{enumerate}[label=(\roman*)]
\item Illustrate the possible outcomes for the 1000,1100 and 1200 trains on a probability tree diagram.
\item Find the probability that\\
(A) all three of these trains are on time,\\
(B) just one of these three trains is on time,\\
(C) the 1200 train is on time.
\item Given that the 1200 train is on time, find the probability that the 1000 train is also on time.
3
\item Write any calculations on page 5.\\
\includegraphics[max width=\textwidth, alt={}, center]{091d6f43-ad01-4849-9f3c-3e58349aa169-4_2276_1490_324_363}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2010 Q7 [18]}}