| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Specific items together |
| Difficulty | Moderate -0.8 This is a straightforward permutations question requiring basic counting principles. Part (i) uses the standard 'treat as a block' method (7!/8! = 2/8). Parts (ii) and (iii) involve simple case analysis with small numbers (4 positions each), requiring only elementary probability calculations without complex reasoning or novel insight. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2 \times 7!}{8!}\) | M1 | \(7!\) and \(8!\) used or \(^7P_7\) and \(^8P_8\) |
| M1 | Correct formula with "\(2 \times\)" | |
| \(= \frac{1}{4}\) | A1 | 3 marks, answer \(\frac{1}{4}\) or \(0.25\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{4}\) or \(4! \times 4!\) or \(3! \times 3!\) or \(\frac{3!}{4!}\) | M1 | |
| \(\left(\frac{1}{4}\right)^2\) or \(\frac{3! \times 3!}{4! \times 4!}\) | M1 | Correct expression |
| \(= \frac{1}{16}\) | A1 | 3 marks, or \(0.0625\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt subdivide, allow one error | M1 | By description or listing or implied by probs, e.g. \(1 - \text{(ii)} - P(\text{sep by 1})\) |
| Correct subdivision into 3 or 13 cases | M1 | All 3 or all 13 cases clearly present |
| Correct expression | M1 | |
| \(= \frac{13}{16}\) | A1 | 4 marks, or \(0.8125\) or a.r.t. \(0.813\) only |
# Question 8:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2 \times 7!}{8!}$ | M1 | $7!$ and $8!$ used or $^7P_7$ and $^8P_8$ |
| | M1 | Correct formula with "$2 \times$" |
| $= \frac{1}{4}$ | A1 | 3 marks, answer $\frac{1}{4}$ or $0.25$ only |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{4}$ or $4! \times 4!$ or $3! \times 3!$ or $\frac{3!}{4!}$ | M1 | |
| $\left(\frac{1}{4}\right)^2$ or $\frac{3! \times 3!}{4! \times 4!}$ | M1 | Correct expression |
| $= \frac{1}{16}$ | A1 | 3 marks, or $0.0625$ |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt subdivide, allow one error | M1 | By description or listing or implied by probs, e.g. $1 - \text{(ii)} - P(\text{sep by 1})$ |
| Correct subdivision into 3 or 13 cases | M1 | All 3 or all 13 cases clearly present |
| Correct expression | M1 | |
| $= \frac{13}{16}$ | A1 | 4 marks, or $0.8125$ or a.r.t. $0.813$ only |
*Eg correct:* $1 - 3 \times \frac{1}{16}\ ;\ 1 - \text{(ii)} - 2 \times \frac{3! \times 3!}{4! \times 4!}$
$\frac{3! \times 3! \times 13}{(4! \times 4!)}\ ;\ \left(\frac{3}{4}\right)^2 + 2 \times \frac{1}{4} \times \frac{2}{4}$
*Eg incorrect:* $1 - \frac{3! \times 3! \times 3}{8!}$ : M1M1M0A0; $1 - \frac{1}{16} - \frac{3! \times 3!}{4! \times 4!}$ : M1M0M0A0
---8 An examination paper consists of 8 questions, of which one is on geometric distributions and one is on binomial distributions.\\
(i) If the 8 questions are arranged in a random order, find the probability that the question on geometric distributions is next to the question on binomial distributions.
Four of the questions, including the one on geometric distributions, are worth 7 marks each, and the remaining four questions, including the one on binomial distributions, are worth 9 marks each. The 7-mark questions are the first four questions on the paper, but are arranged in random order. The 9-mark questions are the last four questions, but are arranged in random order. Find the probability that\\
(ii) the questions on geometric distributions and on binomial distributions are next to one another,\\
(iii) the questions on geometric distributions and on binomial distributions are separated by at least 2 other questions.
\hfill \mbox{\textit{OCR S1 2005 Q8 [10]}}