| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Correct ordering probability |
| Difficulty | Easy -1.3 This is a straightforward S1 probability question testing basic counting principles. Part (i) requires recognizing there are 5! arrangements with only 1 favorable outcome (1/120). Part (ii) is a simple combination calculation: C(2,5) = 10 possible pairs with 1 favorable outcome (1/10). Both parts are routine applications of fundamental counting with no problem-solving insight required, making this easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks |
|---|---|
| Marks: M1 for 5! or 120 or *P\(_5\): seen or product of correct fractions, A1 CAO | 2 |
| Answer | Marks |
|---|---|
| Marks: M1 for denominators, M1 for \(2 \times\) dep on correct denominators, A1 CAO; Or M1 for \(\left(\frac{5}{2}\right)\) or 10, M1 for \(1/\left(\frac{5}{2}\right)\) | 3 |
## (i)
**Answer/Working:** Either $P(\text{alphabetic order}) = \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1} = \frac{1}{120}$ or $P(\text{alphabetic order}) = \frac{1}{5!} = \frac{1}{120} = 0.00833$
**Marks:** M1 for 5! or 120 or *P$_5$: seen or product of correct fractions, A1 CAO | 2
**Guidance:** Allow 0.0083 or 1/120 but not 0.008
## (ii)
**Answer/Working:** Either $P(\text{picks Austen and Bronte}) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ or $P(\text{picks Austen and Bronte}) = \frac{1}{5} \times \frac{1}{4} \times 2 = \frac{1}{10}$ or $P(\text{picks Austen and Bronte}) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$ or M1 for $\left(\frac{5}{2}\right)$ or 10, M1 for $\left(\frac{5}{2}\right)$
**Marks:** M1 for denominators, M1 for $2 \times$ dep on correct denominators, A1 CAO; Or M1 for $\left(\frac{5}{2}\right)$ or 10, M1 for $1/\left(\frac{5}{2}\right)$ | 3
**Guidance:** 1/$P_2$ scores M1 also 1/20 oe scores M1 even if followed by further incorrect working. $\left(\frac{5}{2}\right)$ seen as part of a binomial expression gets M1. Division by 6 or other spurious factor gets MAX M1A0
---I have 5 books, each by a different author. The authors are Austen, Brontë, Clarke, Dickens and Eliot.
\begin{enumerate}[label=(\roman*)]
\item If I arrange the books in a random order on my bookshelf, find the probability that the authors are in alphabetical order with Austen on the left. [2]
\item If I choose two of the books at random, find the probability that I choose the books written by Austen and Brontë. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2011 Q2 [5]}}