| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Correct ordering probability |
| Difficulty | Easy -1.3 This is a straightforward permutations question requiring only basic counting principles and probability calculations. Part (i) is direct factorial application (5!), parts (ii-iii) involve simple favorable outcomes over total outcomes, and part (iv) requires recognizing 3! arrangements. No problem-solving insight needed—purely routine recall and application of standard formulas from early statistics/counting topics. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Number of ways \(= 5! = 120\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probability \(= \frac{2}{120}\) | M1 | For division by their 120; M1 for \(\frac{k}{120}\) |
| \(= \frac{1}{60}\) or \(0.0167\) or \(0.01\dot{6}\) | A1 [2] | CAO; Condone final answer of 2/120; Do not allow 0.016 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{60}\) or \(0.0167\) or \(0.01\dot{6}\) | B1 [1] | Condone 2/120 for B1; Do not allow 0.016 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{1}{10}\) | M1 | For \(\frac{3}{5}\times\); Listing options gives 12/120 |
| A1 | CAO; Or \((3!/120)\times 2\); Or \(\binom{3}{P_3} \times \binom{2}{P_2}/120\) | |
| or \(\frac{1}{\binom{5}{C_3}} = \frac{1}{10}\) | M1 A1 [2] | For division by \(^5C_3\); SC2 for \(3! \times\) their part (iii) or \(6\times\) their part (iii) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Number of ways $= 5! = 120$ | B1 [1] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probability $= \frac{2}{120}$ | M1 | For division by their 120; M1 for $\frac{k}{120}$ |
| $= \frac{1}{60}$ or $0.0167$ or $0.01\dot{6}$ | A1 [2] | CAO; Condone final answer of 2/120; Do not allow 0.016 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{60}$ or $0.0167$ or $0.01\dot{6}$ | B1 [1] | Condone 2/120 for B1; Do not allow 0.016 |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{1}{10}$ | M1 | For $\frac{3}{5}\times$; Listing options gives 12/120 |
| | A1 | CAO; Or $(3!/120)\times 2$; Or $\binom{3}{P_3} \times \binom{2}{P_2}/120$ |
| or $\frac{1}{\binom{5}{C_3}} = \frac{1}{10}$ | M1 A1 [2] | For division by $^5C_3$; SC2 for $3! \times$ their part (iii) or $6\times$ their part (iii) |
---3 (i) There are 5 runners in a race. How many different finishing orders are possible? [You should assume that there are no 'dead heats', where two runners are given the same position.]
For the remainder of this question you should assume that all finishing orders are equally likely.\\
(ii) The runners are denoted by $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E }$. Find the probability that they either finish in the order ABCDE or in the order EDCBA.\\
(iii) Find the probability that the first 3 runners to finish are $\mathrm { A } , \mathrm { B }$ and C , in that order.\\
(iv) Find the probability that the first 3 runners to finish are $\mathrm { A } , \mathrm { B }$ and C , in any order.
\hfill \mbox{\textit{OCR MEI S1 2016 Q3 [6]}}