| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Independent repeated trials |
| Difficulty | Moderate -0.8 This is a straightforward application of independent probability rules with clear structure. Part (i) requires simple multiplication of probabilities, part (ii) uses complement or addition rule, and part (iii) applies conditional probability formula. All calculations are routine with no conceptual challenges beyond basic independence concepts. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Probability \(= 0.3 \times 0.8 = 0.24\) | M1 | for \(0.8\) from \((1-0.2)\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) | M1 | for adding \(0.3\) and \(0.2\) |
| \(= 0.3 + 0.2 - 0.3 \times 0.2\) | M1 | for subtraction of \((0.3 \times 0.2)\) |
| \(= 0.5 - 0.06 = 0.44\) | A1 | cao |
| Or: \(P(A \cup B) = 0.7 \times 0.2 + 0.3 \times 0.8 + 0.3 \times 0.2 = 0.44\) | M1, M1, A1 | M1 either of first terms; M1 for last term |
| Or: \(P(A \cup B) = 1 - P(A' \cap B') = 1 - 0.7 \times 0.8 = 0.44\) | M1, M1, A1 | M1 for \(0.7 \times 0.8\) or \(0.56\); M1 for complete method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.06}{0.44} = \frac{6}{44} = 0.136\) | M1 |
| M1 | for 'their \(0.44\)' in denominator | |
| A1 | FT (must be valid \(p\)) |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Probability $= 0.3 \times 0.8 = 0.24$ | M1 | for $0.8$ from $(1-0.2)$ |
| | A1 | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | M1 | for adding $0.3$ and $0.2$ |
| $= 0.3 + 0.2 - 0.3 \times 0.2$ | M1 | for **subtraction** of $(0.3 \times 0.2)$ |
| $= 0.5 - 0.06 = 0.44$ | A1 | cao |
| **Or:** $P(A \cup B) = 0.7 \times 0.2 + 0.3 \times 0.8 + 0.3 \times 0.2 = 0.44$ | M1, M1, A1 | M1 either of first terms; M1 for last term |
| **Or:** $P(A \cup B) = 1 - P(A' \cap B') = 1 - 0.7 \times 0.8 = 0.44$ | M1, M1, A1 | M1 for $0.7 \times 0.8$ or $0.56$; M1 for complete method |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.06}{0.44} = \frac{6}{44} = 0.136$ | M1 | for numerator of their $0.06$ only |
| | M1 | for 'their $0.44$' in denominator |
| | A1 | FT (must be valid $p$) |
---2 Steve is going on holiday. The probability that he is delayed on his outward flight is 0.3 . The probability that he is delayed on his return flight is 0.2 , independently of whether or not he is delayed on the outward flight.\\
(i) Find the probability that Steve is delayed on his outward flight but not on his return flight.\\
(ii) Find the probability that he is delayed on at least one of the two flights.\\
(iii) Given that he is delayed on at least one flight, find the probability that he is delayed on both flights.
\hfill \mbox{\textit{OCR MEI S1 Q2 [8]}}