| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Moderate -0.3 This is a standard S1 combinations question with three straightforward parts: (i) uses basic conditional probability with combinations, (ii) applies standard 'treat as unit' technique with one person fixed, (iii) extends this to a larger unit. All parts follow textbook methods with no novel insight required, making it slightly easier than average but not trivial due to the multi-part nature and need for careful counting. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^5C_2\) oe seen anywhere or num \(= 10\) alone | M1 | \(\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\) or \(\frac{20}{1680}\) or \(\frac{1}{84}\) oe seen — alone or \(\times\ldots\) eg \(\frac{2}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\) M1 |
| \(\frac{^5C_2}{^8C_4}\) oe or \(\frac{^5C_2\times4!}{^8P_4}\) oe all correct | M1 | \(\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times ^4C_2\times 2\) or \(\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times4!\div 2\) oe or \(\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times 12\) oe all correct — \(\frac{4}{8}\times\frac{3}{7}\times\frac{4}{6}\) oe all correct M2. NB \(\frac{\text{incorrect}}{^8C_4}\) does not score |
| \(= \frac{1}{7}\) or \(0.143\) (3 sf) | A1 | Correct ans scores M1M1A1 regardless of method |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6!\times 2\) alone or \(5!\times 6\times 2\) alone oe | M2 | M1 for \(6!\) or \(5!\times 6\) or \(^6P_5\) or 720 seen. NB 5! scores M0 unless \(5!\times 6\) or \(5!\times 12\) — M1 for \(7!\times 2\) alone. NB \(7!\) scores M0 unless \(7!\times 2\) alone |
| \(= 1440\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6!\times 4\) alone or \(6!\times 2\times 2\) alone | M2 | M1 for \(6!\) or \(^6P_5\) or 720 seen or \(5!\times 6\) seen but NOT from \(5!\times3!\) — 5!: M0 unless \(5!\times6\) or \(5!\times12\) or \(5!\times24\) |
| \(= 2880\) | A1 | |
| [3] |
# Question 8:
## Part 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^5C_2$ oe seen anywhere or num $= 10$ alone | M1 | $\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}$ or $\frac{20}{1680}$ or $\frac{1}{84}$ oe seen — alone or $\times\ldots$ eg $\frac{2}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}$ M1 |
| $\frac{^5C_2}{^8C_4}$ oe or $\frac{^5C_2\times4!}{^8P_4}$ oe all correct | M1 | $\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times ^4C_2\times 2$ or $\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times4!\div 2$ oe or $\frac{1}{8}\times\frac{1}{7}\times\frac{5}{6}\times\frac{4}{5}\times 12$ oe all correct — $\frac{4}{8}\times\frac{3}{7}\times\frac{4}{6}$ oe all correct M2. NB $\frac{\text{incorrect}}{^8C_4}$ does not score |
| $= \frac{1}{7}$ or $0.143$ (3 sf) | A1 | Correct ans scores M1M1A1 regardless of method |
| **[3]** | | |
## Part 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6!\times 2$ alone or $5!\times 6\times 2$ alone oe | M2 | M1 for $6!$ or $5!\times 6$ or $^6P_5$ or 720 seen. NB 5! scores M0 unless $5!\times 6$ or $5!\times 12$ — M1 for $7!\times 2$ alone. NB $7!$ scores M0 unless $7!\times 2$ alone |
| $= 1440$ | A1 | |
| **[3]** | | |
## Part 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6!\times 4$ alone or $6!\times 2\times 2$ alone | M2 | M1 for $6!$ or $^6P_5$ or 720 seen or $5!\times 6$ seen but NOT from $5!\times3!$ — 5!: M0 unless $5!\times6$ or $5!\times12$ or $5!\times24$ |
| $= 2880$ | A1 | |
| **[3]** | | |
---8 A group of 8 people, including Kathy, David and Harpreet, are planning a theatre trip.\\
(i) Four of the group are chosen at random, without regard to order, to carry the refreshments. Find the probability that these 4 people include Kathy and David but not Harpreet.\\
(ii) The 8 people sit in a row. Kathy and David sit next to each other and Harpreet sits at the left-hand end of the row. How many different arrangements of the 8 people are possible?\\
(iii) The 8 people stand in a line to queue for the exit. Kathy and David stand next to each other and Harpreet stands next to them. How many different arrangements of the 8 people are possible?
\hfill \mbox{\textit{OCR S1 2014 Q8 [9]}}