| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | P(X ≤ n) or P(X < n) |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution formula with direct substitution. Part (i) requires P(X=4) = (1/3)³(2/3), part (ii) needs summing three terms or using 1-P(X≥4), and part (iii) is the standard expectation formula E(X)=1/p=3/2. All parts are routine recall and basic calculation with no problem-solving insight required, making it easier than average. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Geo(\(\frac{2}{5}\)) stated; \((\frac{1}{3})^x \frac{2}{3}\) | M1 | or implied by \((\frac{1}{3})^n \times \frac{2}{3}\) |
| M1 | ||
| \(= \frac{2}{81}\) or \(0.0247\) (3 sfs) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{(\frac{1}{3})^3}{1 - (\frac{1}{3})^3}\) | M1 | or \(\frac{2}{3} + \frac{1}{3} x^2 + \frac{1}{3} (\frac{1}{3})^2 x^3 + ((\frac{1}{3})x)^2\) or M2; one term omitted or extra or wrong: M1 |
| M1 | ||
| \(\frac{26}{27}\) or \(0.963\) (3 sfs) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{1 - \frac{2}{3}}\) | M1 | |
| \(= \frac{3}{2}\) oe | A1 | 2 marks |
**i)**
Geo($\frac{2}{5}$) stated; $(\frac{1}{3})^x \frac{2}{3}$ | M1 | or implied by $(\frac{1}{3})^n \times \frac{2}{3}$
| M1 |
$= \frac{2}{81}$ or $0.0247$ (3 sfs) | A1 | 3 marks
**Total for Question 6:** 3 marks
## Question 6ii:
$\frac{(\frac{1}{3})^3}{1 - (\frac{1}{3})^3}$ | M1 | or $\frac{2}{3} + \frac{1}{3} x^2 + \frac{1}{3} (\frac{1}{3})^2 x^3 + ((\frac{1}{3})x)^2$ or M2; one term omitted or extra or wrong: M1
| M1 |
$\frac{26}{27}$ or $0.963$ (3 sfs) | A1 | 3 marks
**iii)**
$\frac{1}{1 - \frac{2}{3}}$ | M1 |
$= \frac{3}{2}$ oe | A1 | 2 marks
**Total for Question 6ii, iii:** 8 marks6 A coin is biased so that the probability that it will show heads on any throw is $\frac { 2 } { 3 }$. The coin is thrown repeatedly.
The number of throws up to and including the first head is denoted by $X$. Find\\
(i) $\mathrm { P } ( X = 4 )$,\\
(ii) $\mathrm { P } ( X < 4 )$,\\
(iii) $\mathrm { E } ( X )$.
\hfill \mbox{\textit{OCR S1 2007 Q6 [8]}}