| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Three or more stages |
| Difficulty | Moderate -0.3 This is a standard three-stage tree diagram problem with conditional probability. While it requires multiple steps including drawing the tree, calculating various probabilities, and applying Bayes' theorem in part (iii), all techniques are routine for S1. The conditional probabilities are clearly stated, and the question follows a typical textbook structure with no novel insight required. |
| 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 |
|---|---|---|
| Tree diagram with three sets of branches, probabilities 0.95 (On time) and 0.05 (Late) at each stage, with 0.6/0.4 splits on Late branches | G1 first set of branches; G1 indep second set of branches; G1 indep third set of branches; G1 labels | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{all on time}) = 0.95^3 = 0.8574\) | M1 for \(0.95^3\); A1 CAO | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{just one on time}) = 0.95 \times 0.05 \times 0.4 + 0.05 \times 0.6 \times 0.05 + 0.05 \times 0.4 \times 0.6 = 0.019 + 0.0015 + 0.012 = 0.0325\) | M1 first term; M1 second term; M1 third term; A1 CAO | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{1200 on time}) = 0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6 + 0.05 \times 0.6 \times 0.95 + 0.05 \times 0.4 \times 0.6 = 0.857375 + 0.0285 + 0.0285 + 0.012 = 0.926375\) | M1 any two terms; M1 third term; M1 fourth term; A1 CAO | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{1000 on time} \mid \text{1200 on time}) = \frac{P(\text{1000 on time and 1200 on time})}{P(\text{1200 on time})} = \frac{0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6}{0.926375} = \frac{0.885875}{0.926375} = 0.9563\) | M1 either term of numerator; M1 full numerator; M1 denominator; A1 CAO | 4 marks |
## Question 3:
### Part (i)
Tree diagram with three sets of branches, probabilities 0.95 (On time) and 0.05 (Late) at each stage, with 0.6/0.4 splits on Late branches | G1 first set of branches; G1 indep second set of branches; G1 indep third set of branches; G1 labels | **4 marks**
### Part (ii)(A)
$P(\text{all on time}) = 0.95^3 = 0.8574$ | M1 for $0.95^3$; A1 CAO | **2 marks**
### Part (ii)(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**
### Part (ii)(C)
$P(\text{1200 on time}) = 0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6 + 0.05 \times 0.6 \times 0.95 + 0.05 \times 0.4 \times 0.6 = 0.857375 + 0.0285 + 0.0285 + 0.012 = 0.926375$ | M1 any two terms; M1 third term; M1 fourth term; A1 CAO | **4 marks**
### Part (iii)
$P(\text{1000 on time} \mid \text{1200 on time}) = \frac{P(\text{1000 on time and 1200 on time})}{P(\text{1200 on time})} = \frac{0.95 \times 0.95 \times 0.95 + 0.95 \times 0.05 \times 0.6}{0.926375} = \frac{0.885875}{0.926375} = 0.9563$ | M1 either term of numerator; M1 full numerator; M1 denominator; A1 CAO | **4 marks**
---3 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.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q3 [18]}}