| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in lines |
| Difficulty | Moderate -0.8 Part (a) is a straightforward probability calculation using basic counting principles (2/8 × 1/7). Part (b) requires identifying cases ('more women than men' means 2 or 3 women) and computing combinations, but this is standard AS-level material with no conceptual challenges. Both parts are routine applications of fundamental probability concepts. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{8} \times \frac{1}{7}\) | M1 (1.1a) | or \(\frac{6!}{8!}\) oe |
| \(\frac{1}{56}\) | A1 (1.1) | CAO accept 0.0179; 0.017857142 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^5C_3\) or \(^5C_2 \times {^3C_1}\) seen \([10, 30]\) | M1 (1.1a) | OR 3W: 10 ways; 2W1M: \(10\times3=30\) ways; 1W2M: \(5\times3=15\) ways; 3M: 1 way. Allow M1 if the 2 cases are identified and added, whatever the calculation. Allow method by subtraction from 1. This question can be done by tree diagram, with similar working. |
| \(^8C_3\) is denominator \([56]\) \(\{\text{equiv to } ^8C_5\}\) | B1 (1.1) | |
| \(P(3,0) + P(2,1)\) attempted | M1 (2.1) | So \(\frac{10+30}{10+30+15+1} = \frac{5}{7}\) |
| \(\frac{5}{7}\) | A1 (1.1) | |
| [4] | ||
| alternatively \(\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}\) or \(3 \times \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}\) seen | M1 (1.1a) | \(\frac{60}{336}\) or \(\frac{5}{28}\) or \(\frac{180}{336}\) or \(\frac{15}{28}\). Allow both M marks for all denoms \(8\): e.g. \(\frac{5}{8} \times \frac{4}{8} \times \frac{3}{8} = \frac{60}{512}\) etc leading to \(\frac{15}{32}\) |
| both correct including \(\times 3\) | A1 (1.1) | |
| \(P(3,0) + P(2,1)\) attempted | M1 (2.1) | Allow even if \(\times3\) missing |
| \(\frac{5}{7}\) | A1 (1.1) | |
| [4] |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{8} \times \frac{1}{7}$ | M1 (1.1a) | or $\frac{6!}{8!}$ oe |
| $\frac{1}{56}$ | A1 (1.1) | CAO accept 0.0179; 0.017857142 |
| **[2]** | | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^5C_3$ or $^5C_2 \times {^3C_1}$ seen $[10, 30]$ | M1 (1.1a) | OR 3W: 10 ways; 2W1M: $10\times3=30$ ways; 1W2M: $5\times3=15$ ways; 3M: 1 way. Allow M1 if the 2 cases are identified and added, whatever the calculation. Allow method by subtraction from 1. This question can be done by tree diagram, with similar working. |
| $^8C_3$ is denominator $[56]$ $\{\text{equiv to } ^8C_5\}$ | B1 (1.1) | |
| $P(3,0) + P(2,1)$ attempted | M1 (2.1) | So $\frac{10+30}{10+30+15+1} = \frac{5}{7}$ |
| $\frac{5}{7}$ | A1 (1.1) | |
| **[4]** | | |
| alternatively $\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}$ or $3 \times \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}$ seen | M1 (1.1a) | $\frac{60}{336}$ or $\frac{5}{28}$ or $\frac{180}{336}$ or $\frac{15}{28}$. Allow both M marks for all denoms $8$: e.g. $\frac{5}{8} \times \frac{4}{8} \times \frac{3}{8} = \frac{60}{512}$ etc leading to $\frac{15}{32}$ |
| both correct including $\times 3$ | A1 (1.1) | |
| $P(3,0) + P(2,1)$ attempted | M1 (2.1) | Allow even if $\times3$ missing |
| $\frac{5}{7}$ | A1 (1.1) | |
| **[4]** | | |
---5 A company needs to appoint 3 new assistants. 8 candidates are invited for interview; each candidate has a different surname. The candidates are to be interviewed one after another. The personnel officer randomly selects the order in which the candidates are to be interviewed by drawing their names out of a hat. One of the candidates is called Mr Browne and another is called Mrs Green.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that Mr Browne is interviewed first and Mrs Green is interviewed last.
5 of the 8 candidates invited for interview are women and the other 3 are men. The chief executive can't make up his mind who to appoint so he randomly selects 3 candidates by drawing their names out of a hat.
\item Determine the probability that more women than men are selected.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2020 Q5 [6]}}