| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties with clear structure: part (i) is trivial scaling, parts (ii)-(iii) require basic probability calculations with independent events and time-scaling, and part (iv) involves summing two Poisson distributions. All steps are standard textbook exercises requiring no novel insight, though the multi-part nature and need to handle complements/sums of independent Poissons places it slightly above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\lambda = 1.15\) | B1 [1] | |
| (ii) \(P(0) \times P(>0) = c^{-1.15} \times (1 - e^{-1.15}) = 0.3166 \times 0.6833 = 0.216\) | M1 | Multiplying two Poisson probs meant to be no goals in first half and something in second half |
| A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{at least 1}) = 1 - P(0) = 1 - e^{-1.533} = 0.784\) | B1 | New mean |
| M1 | Attempt at finding \(1 - P(0)\) with new mean | |
| A1 [3] | Correct answer (cwo) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{at least 3}) = 1 - P(0, 1, 2) = 1 - e^{-4.1}\left(1 + 4.1 + \frac{4.1^2}{2}\right) = 1 - 0.224 = 0.776\) | B1 | New mean (or 6 correct combinations 0,0, 1,0, 2,0 etc) |
| M1 | Using Poisson with new mean (or combinations) to find P (≥ 3) condone 1 end error | |
| A1 [3] | Correct answer |
(i) $\lambda = 1.15$ | B1 [1] |
(ii) $P(0) \times P(>0) = c^{-1.15} \times (1 - e^{-1.15}) = 0.3166 \times 0.6833 = 0.216$ | M1 | Multiplying two Poisson probs meant to be no goals in first half and something in second half
| A1 [2] | Correct answer
(iii) $\lambda = \frac{60}{90} \times 2.3 = 1.53(3)$
$P(\text{at least 1}) = 1 - P(0) = 1 - e^{-1.533} = 0.784$ | B1 | New mean
| M1 | Attempt at finding $1 - P(0)$ with new mean
| A1 [3] | Correct answer (cwo)
(iv) $\lambda = 4.1$
$P(\text{at least 3}) = 1 - P(0, 1, 2) = 1 - e^{-4.1}\left(1 + 4.1 + \frac{4.1^2}{2}\right) = 1 - 0.224 = 0.776$ | B1 | New mean (or 6 correct combinations 0,0, 1,0, 2,0 etc)
| M1 | Using Poisson with new mean (or combinations) to find P (≥ 3) condone 1 end error
| A1 [3] | Correct answer
---6 In their football matches, Rovers score goals independently and at random times. Their average rate of scoring is 2.3 goals per match.\\
(i) State the expected number of goals that Rovers will score in the first half of a match.\\
(ii) Find the probability that Rovers will not score any goals in the first half of a match but will score one or more goals in the second half of the match.\\
(iii) Football matches last for 90 minutes. In a particular match, Rovers score one goal in the first 30 minutes. Find the probability that they will score at least one further goal in the remaining 60 minutes.
Independently of the number of goals scored by Rovers, the number of goals scored per football match by United has a Poisson distribution with mean 1.8.\\
(iv) Find the probability that a total of at least 3 goals will be scored in a particular match when Rovers play United.
\hfill \mbox{\textit{CAIE S2 2008 Q6 [9]}}