| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Mutually exclusive event categories |
| Difficulty | Easy -1.2 This is a straightforward tree diagram question testing basic probability rules (complements, multiplication, addition) with clearly stated probabilities and standard multi-part structure. All parts follow routine procedures with no problem-solving insight required—significantly easier than average A-level questions. |
| 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 |
|---|---|---|
| \(a = 0.8,\ b = 0.85,\ c = 0.9\) | B1 for any one, B1 for the other two | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Not delayed}) = 0.8 \times 0.85 \times 0.9 = 0.612\) | M1, A1 CAO | |
| \(P(\text{Delayed}) = 1 - 0.8 \times 0.85 \times 0.9 = 1 - 0.612 = 0.388\) | M1 for \(1 - P(\text{delayed})\), A1FT | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.153 + 0.108 + 0.068 = 0.329\) | B1 one product correct, M1 three products, M1 sum of 3 products, A1 CAO | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{just one problem} \mid \text{delay}) = \frac{P(\text{just one problem and delay})}{P(\text{delay})} = \frac{0.329}{0.388} = 0.848\) | M1 for numerator, M1 for denominator, A1FT | 3 marks |
| Answer | Marks |
|---|---|
| M1 for \(0.15+\), M1 for second term, A1CAO / M1 for product, M1 for \(1-\text{product}\), A1CAO / M1 for all 3 products, M1 for sum, A1CAO / M1 numerator, M1 denominator, A1CAO | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Expected number \(= 110 \times 0.388 = 42.7\) | M1 for product, A1FT | Total: 18 marks |
# Question 7:
## Part (i):
$a = 0.8,\ b = 0.85,\ c = 0.9$ | B1 for any one, B1 for the other two | **2 marks**
## Part (ii):
$P(\text{Not delayed}) = 0.8 \times 0.85 \times 0.9 = 0.612$ | M1, A1 CAO |
$P(\text{Delayed}) = 1 - 0.8 \times 0.85 \times 0.9 = 1 - 0.612 = 0.388$ | M1 for $1 - P(\text{delayed})$, A1FT | **4 marks**
## Part (iii):
$P(\text{just one problem})$
$= 0.2\times0.85\times0.9 + 0.8\times0.15\times0.9 + 0.8\times0.85\times0.1$
$= 0.153 + 0.108 + 0.068 = 0.329$ | B1 one product correct, M1 three products, M1 sum of 3 products, A1 CAO | **4 marks**
## Part (iv):
$P(\text{just one problem} \mid \text{delay}) = \frac{P(\text{just one problem and delay})}{P(\text{delay})} = \frac{0.329}{0.388} = 0.848$ | M1 for numerator, M1 for denominator, A1FT | **3 marks**
## Part (v):
$P(\text{Delayed} \mid \text{No technical problems})$
Either $= 0.15 + 0.85 \times 0.1 = 0.235$
Or $= 1 - 0.9 \times 0.85 = 1 - 0.765 = 0.235$
Or $= 0.15\times0.1 + 0.15\times0.9 + 0.85\times0.1 = 0.235$
Or using conditional probability formula: $\frac{0.8\times0.15\times0.1+0.8\times0.15\times0.9+0.8\times0.85\times0.1}{0.8} = \frac{0.188}{0.8} = 0.235$
| M1 for $0.15+$, M1 for second term, A1CAO / M1 for product, M1 for $1-\text{product}$, A1CAO / M1 for all 3 products, M1 for sum, A1CAO / M1 numerator, M1 denominator, A1CAO | **3 marks**
## Part (vi):
Expected number $= 110 \times 0.388 = 42.7$ | M1 for product, A1FT | **Total: 18 marks**
---7 Laura frequently flies to business meetings and often finds that her flights are delayed. A flight may be delayed due to technical problems, weather problems or congestion problems, with probabilities $0.2,0.15$ and 0.1 respectively. The tree diagram shows this information.\\
\includegraphics[max width=\textwidth, alt={}, center]{3a5d18f5-b1fe-4513-ae4e-f37c20f172b5-4_608_1651_532_248}\\
(i) Write down the values of the probabilities $a , b$ and $c$ shown in the tree diagram.
One of Laura's flights is selected at random.\\
(ii) Find the probability that Laura's flight is not delayed and hence write down the probability that it is delayed.\\
(iii) Find the probability that Laura's flight is delayed due to just one of the three problems.\\
(iv) Given that Laura's flight is delayed, find the probability that the delay is due to just one of the three problems.\\
(v) Given that Laura's flight has no technical problems, find the probability that it is delayed.\\
(vi) In a particular year, Laura has 110 flights. Find the expected number of flights that are delayed.
\hfill \mbox{\textit{OCR MEI S1 2009 Q7 [18]}}