Edexcel S2 2014 January — Question 5 12 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2014
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson with binomial combination
DifficultyStandard +0.8 This question requires multiple Poisson calculations with rate adjustments (annual to monthly), then crucially applies binomial distribution to model repeated monthly trials in parts (c) and (d). The nested probability structure (Poisson within binomial) and the complementary probability reasoning needed for part (d) elevate this beyond routine S2 questions, though the individual calculations are standard once the structure is identified.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling

5. A school photocopier breaks down randomly at a rate of 15 times per year.
  1. Find the probability that there will be exactly 3 breakdowns in the next month.
  2. Show that the probability that there will be at least 2 breakdowns in the next month is 0.355 to 3 decimal places.
  3. Find the probability of at least 2 breakdowns in each of the next 4 months. The teachers would like a new photocopier. The head teacher agrees to monitor the situation for the next 12 months. The head teacher decides he will buy a new photocopier if there is more than 1 month when the photocopier has at least 2 breakdowns.
  4. Find the probability that the head teacher will buy a new photocopier.

Question 5:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Notes
Let \(X\) = number of breakdowns per month; \(X \sim \text{Po}\!\left(\frac{15}{12}\right)\)B1 Writing or using Po(1.25)
\(P(X=3) = \frac{e^{-1.25}1.25^3}{3!}\)M1 \(\frac{e^{-\lambda}\lambda^3}{3!}\)
\(= 0.0933\)A1 awrt 0.0933
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Notes
\(P(X \geq 2) = 1 - P(X=0) - P(X=1)\)M1 \(1-P(X=0)-P(X=1)\) or \(1-P(X\leq1)\) with correct expression using their \(\lambda\); condone 0.3554 or better. Answer is given (AG) so must show working
\(= 1 - e^{-1.25}(1+1.25)\)
\(= 0.35536...\)
\(= 0.355\)A1cso
Part (c):
AnswerMarks Guidance
Working/AnswerMarks Notes
\((0.355)^4 = 0.0159\)M1A1 M1: their \([(b)]^4\)
Part (d):
AnswerMarks Guidance
Working/AnswerMarks Notes
\(Y \sim\) number of months the photocopier breaks down at least twice
\(Y \sim B(12,\; 0.355)\)M1A1 M1: identifying Binomial; 1st A1: B(12, their (b))
\(P(Y \geq 2) = 1 - P(Y=0) - P(Y=1)\)dM1 \(1-P(Y=0)-P(Y=1)\) or \(1-P(X\leq1)\); dependent on 1st M1
\(= 1-(1-0.355)^{12} - 12(1-0.355)^{11}(0.355)\)A1 2nd A1: correct expression
\(= 0.961\)A1 awrt 0.961
## Question 5:

### Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| Let $X$ = number of breakdowns per month; $X \sim \text{Po}\!\left(\frac{15}{12}\right)$ | B1 | Writing or using Po(1.25) |
| $P(X=3) = \frac{e^{-1.25}1.25^3}{3!}$ | M1 | $\frac{e^{-\lambda}\lambda^3}{3!}$ |
| $= 0.0933$ | A1 | awrt 0.0933 |

### Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $P(X \geq 2) = 1 - P(X=0) - P(X=1)$ | M1 | $1-P(X=0)-P(X=1)$ or $1-P(X\leq1)$ with correct expression using their $\lambda$; condone 0.3554 or better. **Answer is given (AG) so must show working** |
| $= 1 - e^{-1.25}(1+1.25)$ | | |
| $= 0.35536...$ | | |
| $= 0.355$ | A1cso | |

### Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $(0.355)^4 = 0.0159$ | M1A1 | M1: their $[(b)]^4$ |

### Part (d):
| Working/Answer | Marks | Notes |
|---|---|---|
| $Y \sim$ number of months the photocopier breaks down at least twice | | |
| $Y \sim B(12,\; 0.355)$ | M1A1 | M1: identifying Binomial; 1st A1: B(12, their (b)) |
| $P(Y \geq 2) = 1 - P(Y=0) - P(Y=1)$ | dM1 | $1-P(Y=0)-P(Y=1)$ or $1-P(X\leq1)$; dependent on 1st M1 |
| $= 1-(1-0.355)^{12} - 12(1-0.355)^{11}(0.355)$ | A1 | 2nd A1: correct expression |
| $= 0.961$ | A1 | awrt 0.961 |

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5. A school photocopier breaks down randomly at a rate of 15 times per year.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that there will be exactly 3 breakdowns in the next month.
\item Show that the probability that there will be at least 2 breakdowns in the next month is 0.355 to 3 decimal places.
\item Find the probability of at least 2 breakdowns in each of the next 4 months.

The teachers would like a new photocopier. The head teacher agrees to monitor the situation for the next 12 months. The head teacher decides he will buy a new photocopier if there is more than 1 month when the photocopier has at least 2 breakdowns.
\item Find the probability that the head teacher will buy a new photocopier.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2014 Q5 [12]}}