| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| 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 (complementary events, multiplication rule, conditional probability) with clearly stated probabilities and standard multi-part structure. All parts follow routine procedures with no novel problem-solving required, making it easier than average. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
# Question 1
## (i)
B1 for any one: $a = 0.8$, $b = 0.85$, $c = 0.9$
B1 for the other two
**Total: 2 marks**
## (ii)
M1 for product: $P(\text{Not delayed}) = 0.8 \times 0.85 \times 0.9 = 0.612$
A1 CAO
M1 for $1 - P(\text{delayed})$: $P(\text{Delayed}) = 1 - 0.8 \times 0.85 \times 0.9 = 1 - 0.612 = 0.388$
A1 FT
**Total: 4 marks**
## (iii)
B1 one product correct: $P(\text{just one problem}) = 0.2 \times 0.85 \times 0.9 + 0.8 \times 0.15 \times 0.9 + 0.8 \times 0.85 \times 0.1$
M1 three products
M1 sum of 3 products: $= 0.153 + 0.108 + 0.068 = 0.329$
A1 CAO
**Total: 4 marks**
## (iv)
M1 for numerator
M1 for denominator: $P(\text{Just one problem} \mid \text{delay}) = \frac{P(\text{Just one problem and delay})}{P(\text{Delay})} = \frac{0.329}{0.388} = 0.848$
A1 FT
**Total: 3 marks**
## (v)
Method 1:
M1 for $0.15 +$
M1 for second term: $P(\text{Delayed} \mid \text{No technical problems}) = 0.15 + 0.85 \times 0.1 = 0.235$
A1 CAO
Method 2:
M1 for product
M1 for $1 -$ product: $= 1 - 0.9 \times 0.85 = 1 - 0.765 = 0.235$
A1 CAO
Method 3:
M1 for all 3 products
M1 for sum of all 3 products: $= 0.15 \times 0.1 + 0.15 \times 0.9 + 0.85 \times 0.1 = 0.235$
A1 CAO
Method 4 (using conditional probability formula):
M1 for numerator: $P(\text{Delayed and no technical problems}) = 0.8 \times 0.15 \times 0.1 + 0.8 \times 0.15 \times 0.9 + 0.8 \times 0.85 \times 0.1 = 0.188$
M1 for denominator: $P(\text{No technical problems}) = 0.8$
A1 CAO: $= \frac{0.188}{0.8} = 0.235$
**Total: 3 marks**
## (vi)
M1 for product: Expected number $= 110 \times 0.388 = 42.7$
A1 FT
**Total: 2 marks**
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**TOTAL: 18 marks**1 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]{10679ff3-494d-4f4e-a38a-0832faa91690-1_605_1650_534_284}\\
(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 Q1 [18]}}