CAIE S2 2004 June — Question 6 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2004
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeConditional probability with Poisson
DifficultyStandard +0.8 Part (i) is routine Poisson calculation. Part (ii) requires complement rule but is standard. Part (iii) is conceptually demanding—students must recognize that given a total of 5 planes in an hour, the distribution of planes between two half-hour periods follows a binomial distribution (not Poisson), requiring understanding of conditional probability and the relationship between Poisson processes and uniform distribution of events.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities

6 At a certain airfield planes land at random times at a constant average rate of one every 10 minutes.
  1. Find the probability that exactly 5 planes will land in a period of one hour.
  2. Find the probability that at least 2 planes will land in a period of 16 minutes.
  3. Given that 5 planes landed in an hour, calculate the conditional probability that 1 plane landed in the first half hour and 4 in the second half hour.

AnswerMarks Guidance
(i) \(P(5) = e^{-6} \times \frac{6^5}{5!} = 0.161\)M1 A1 For an attempted Poisson P(5) calculation, any mean; For correct answer
2
(ii) \(P(X \geq 2) = 1 - [P(0) + P(1)] = 1 - e^{-1.6}(1 + 1.6) = 0.475\)B1 M1 A1 For \(\mu = 1.6\), evaluated in a Poisson prob; For \(1 - P(0) - P(1)\) or \(1 - P(0) - P(1) - P(2)\); For correct answer
3
(iii) \(P(1 \text{ then } 45) = \frac{(e^{-3} \times 3) \times (e^{-3} \times \frac{3^4}{4!})}{e^{-6} \times \frac{6^5}{5!}} = 0.156\) or \(\frac{5}{32}\) M1 M1 A1
3
**(i)** $P(5) = e^{-6} \times \frac{6^5}{5!} = 0.161$ | M1 A1 | For an attempted Poisson P(5) calculation, any mean; For correct answer

| | **2** | |

**(ii)** $P(X \geq 2) = 1 - [P(0) + P(1)] = 1 - e^{-1.6}(1 + 1.6) = 0.475$ | B1 M1 A1 | For $\mu = 1.6$, evaluated in a Poisson prob; For $1 - P(0) - P(1)$ or $1 - P(0) - P(1) - P(2)$; For correct answer

| | **3** | |

**(iii)** $P(1 \text{ then } 4 | 5) = \frac{(e^{-3} \times 3) \times (e^{-3} \times \frac{3^4}{4!})}{e^{-6} \times \frac{6^5}{5!}} = 0.156$ or $\frac{5}{32}$ | M1 M1 A1 | For multiplying P(1) by P(4) any (consistent) mean; For dividing by P(5) any mean; For correct answer

| | **3** | |
6 At a certain airfield planes land at random times at a constant average rate of one every 10 minutes.\\
(i) Find the probability that exactly 5 planes will land in a period of one hour.\\
(ii) Find the probability that at least 2 planes will land in a period of 16 minutes.\\
(iii) Given that 5 planes landed in an hour, calculate the conditional probability that 1 plane landed in the first half hour and 4 in the second half hour.

\hfill \mbox{\textit{CAIE S2 2004 Q6 [8]}}