| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single time period probability |
| Difficulty | Standard +0.3 This question combines conditional probability (part a), basic probability with sampling without replacement (part b), and Poisson/binomial distribution application (part c). Part (a) requires a simple probability tree over two weeks. Part (b) is straightforward hypergeometric/basic probability. Part (c) requires setting up an inequality with binomial or Poisson approximation. While multi-part, each component uses standard techniques without requiring novel insight, making it slightly easier than average for a Further Maths Statistics question. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.4\times0.4\) or \(0.6\times0.2\) seen | M1 | Or tree diagram with correct outcomes and probabilities shown; allow \(0.4\times0.4\times0.4\) or \(0.4\times0.6\times0.2\) |
| \(0.4\times0.4 + 0.6\times0.2\) | A1 | |
| \(0.28\) | A1 [3] | Mark the final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{120}\) or \(\frac{5}{120}\) seen | M1 | |
| \(\frac{1}{24}\) or \(0.0416666\ldots\) to 2 or more sf | A1 [2] | B2 for \(0.0416666\ldots\) unsupported; B0 for \(0.042\) unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 -\) their \(\frac{1}{24}\) evaluated | M1 | |
| \(1 - \left(\frac{23}{24}\right)^n > 0.95\) FT | M1 | Allow use of \(=\), \(\geq\) or \(\leq\) |
| \(\left(\frac{23}{24}\right)^n < 0.05\) FT | M1 | Allow use of \(=\) or \(\leq\) |
| \(n = 71\) cao | A1 [4] | \(70.3890\ldots\) unsupported but rounded to 1 or more dp implies M1M1M1; award full marks for 71 unsupported or from trial and improvement |
## Question 8:
### Part (a):
$0.4\times0.4$ or $0.6\times0.2$ seen | **M1** | Or tree diagram with correct outcomes and probabilities shown; allow $0.4\times0.4\times0.4$ or $0.4\times0.6\times0.2$
$0.4\times0.4 + 0.6\times0.2$ | **A1** |
$0.28$ | **A1 [3]** | Mark the final answer
### Part (b):
$\frac{1}{120}$ or $\frac{5}{120}$ seen | **M1** |
$\frac{1}{24}$ or $0.0416666\ldots$ to 2 or more sf | **A1 [2]** | **B2** for $0.0416666\ldots$ unsupported; **B0** for $0.042$ unsupported
### Part (c):
$1 -$ their $\frac{1}{24}$ evaluated | **M1** |
$1 - \left(\frac{23}{24}\right)^n > 0.95$ FT | **M1** | Allow use of $=$, $\geq$ or $\leq$
$\left(\frac{23}{24}\right)^n < 0.05$ FT | **M1** | Allow use of $=$ or $\leq$
$n = 71$ cao | **A1 [4]** | $70.3890\ldots$ unsupported but rounded to 1 or more dp implies **M1M1M1**; award full marks for 71 unsupported or from trial and improvement
---8 A team called "The Educated Guess" enter a weekly quiz. If they win the quiz in a particular week, the probability that they will win the following week is 0.4 , but if they do not win, the probability that they will win the following week is 0.2 .
In week 4 The Educated Guess won the quiz.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that The Educated Guess will win the quiz in week 6.
Every week the same 20 quiz teams, each with 6 members, take part in a quiz. Every member of every team buys a raffle ticket. Five winning tickets are drawn randomly, without replacement. Alf, who is a member of one of the teams, takes part every week.
\item Calculate the probability that, in a randomly chosen week, Alf wins a raffle prize.
\item Find the smallest number of weeks after which it will be $95 \%$ certain that Alf has won at least one raffle prize.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q8 [9]}}