| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | P(a ≤ X ≤ b) range probability |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric distribution formulas with clear context. Parts (i) and (ii) use standard P(X=k) and range probability calculations. Parts (iii) and (iv) require slightly more thought (complementary probability and independence across days) but remain routine for S1 level with no novel problem-solving required. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.7^4\times 0.3\) alone | M1 | |
| \(= 0.0720\) (3 sf) or \(\frac{7203}{100000}\) oe | A1 | allow 0.072 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.7 + 0.7^2 + 0.7^3)\times 0.3\) | M2 | M1 for 1 term omitted, wrong or extra; must add terms, not mult — \((1-0.7^4)-0.3\) or \(0.7599-0.3\) M2; \((1-0.7^4)-\ldots\) or \(1-0.3-\ldots\) M1; \(0.7599-\ldots\) or \(0.7-\ldots\) M1; Just \(1-0.7^4\) or \(1-0.3\): M0; \((1+0.7+0.7^2+0.7^3)\times0.3-0.3\) M2; 1 term omitted, wrong or extra M1 |
| \(= 0.4599\) or \(0.460\) (3sf) or \(\frac{4599}{10000}\) oe | A1 | Allow 0.46 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.7^6\) | M2 | M1 for \(0.7^6\) alone or \(1-0.7^5\ (=0.832)\) or \(1-0.7^7\ (=0.918)\) — \(0.3(1+0.7+0.7^2+0.7^3+0.7^4+0.7^5)\) M2; or (ii) \(+ 0.3(1+0.7^4+0.7^5)\) M2; or (i) \(+\) (ii) \(+ 0.3(1+0.7^5)\) M2; one term omitted or extra: M1; must add terms, not mult. NB ans 0.832 might be M1M0A0 from omitting last term. Could be, eg, their (ii) \(+0.3(1+0.7^4)\). correct working, but subtr from 1: M1 |
| \(= 0.882\) (3 sf) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 - \text{"0.882"})^2 \times \text{"0.882"}\) | M1 | or \((0.7^6)^2 \times (1 - 0.7^6)\) or \(0.1176^2 \times (1 - 0.1176)\) or \((0.7^6)^2 \times \text{their "0.882"}\) or \(0.3(0.7^{12} + 0.7^{13} + 0.7^{14} + \ldots + 0.7^{17})\) |
| \(= 0.0122\) (3 sf) | A1ft | allow \(0.0123\) |
# Question 9:
**Note:** If 0.3 and 0.7 are interchanged consistently through all four parts, all M-marks can be scored, but no A-marks. If $1-0.3$ is calculated incorrectly (eg 0.6 or 0.66 or $\frac{2}{3}$) consistently, lose the A-mark in (i) but all other marks are available on ft, so long as $0 < \text{ans} < 1$.
## Part 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.7^4\times 0.3$ alone | M1 | |
| $= 0.0720$ (3 sf) or $\frac{7203}{100000}$ oe | A1 | allow 0.072 |
| **[2]** | | |
## Part 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.7 + 0.7^2 + 0.7^3)\times 0.3$ | M2 | M1 for 1 term omitted, wrong or extra; must add terms, not mult — $(1-0.7^4)-0.3$ or $0.7599-0.3$ M2; $(1-0.7^4)-\ldots$ or $1-0.3-\ldots$ M1; $0.7599-\ldots$ or $0.7-\ldots$ M1; Just $1-0.7^4$ or $1-0.3$: M0; $(1+0.7+0.7^2+0.7^3)\times0.3-0.3$ M2; 1 term omitted, wrong or extra M1 |
| $= 0.4599$ or $0.460$ (3sf) or $\frac{4599}{10000}$ oe | A1 | Allow 0.46 |
| **[3]** | | |
## Part 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.7^6$ | M2 | M1 for $0.7^6$ alone or $1-0.7^5\ (=0.832)$ or $1-0.7^7\ (=0.918)$ — $0.3(1+0.7+0.7^2+0.7^3+0.7^4+0.7^5)$ M2; or (ii) $+ 0.3(1+0.7^4+0.7^5)$ M2; or (i) $+$ (ii) $+ 0.3(1+0.7^5)$ M2; one term omitted or extra: M1; must add terms, not mult. NB ans 0.832 might be M1M0A0 from omitting last term. Could be, eg, their (ii) $+0.3(1+0.7^4)$. correct working, but subtr from 1: M1 |
| $= 0.882$ (3 sf) | A1 | |
| **[3]** | | |
## Question 9(iv):
$(1 - \text{"0.882"})^2 \times \text{"0.882"}$ | M1 | or $(0.7^6)^2 \times (1 - 0.7^6)$ or $0.1176^2 \times (1 - 0.1176)$ or $(0.7^6)^2 \times \text{their "0.882"}$ or $0.3(0.7^{12} + 0.7^{13} + 0.7^{14} + \ldots + 0.7^{17})$ | Not $0.7^2 \times 0.3$
$= 0.0122$ (3 sf) | A1ft | allow $0.0123$ | Completely correct method; ft their "0.882" except if 0.3 or 0.7
**Total: [2]**9 Each day Harry makes repeated attempts to light his gas fire. If the fire lights he makes no more attempts. On each attempt, the probability that the fire will light is 0.3 independent of all other attempts. Find the probability that\\
(i) the fire lights on the 5th attempt,\\
(ii) Harry needs more than 1 attempt but fewer than 5 attempts to light the fire.
If the fire does not light on the 6th attempt, Harry stops and the fire remains unlit.\\
(iii) Find the probability that, on a particular day, the fire lights.\\
(iv) Harry's week starts on Monday. Find the probability that, during a certain week, the first day on which the fire lights is Wednesday.
\hfill \mbox{\textit{OCR S1 2014 Q9 [10]}}