| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | P(a ≤ X ≤ b) range probability |
| Difficulty | Moderate -0.8 This is a straightforward geometric distribution question requiring only direct application of the formula P(X=k) = (1-p)^(k-1) × p with p=0.3. All three parts involve routine calculations with no conceptual challenges—part (i) is direct substitution, part (ii) uses the complement rule or geometric series, and part (iii) sums probabilities. This is easier than average as it tests only basic probability distribution knowledge with clear setup and standard techniques. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
### Part i
**Answer:** Geo stated $0.7^3 \times 0.3$ alone = $\frac{1029}{10000}$ or 0.103 (3 sf)
**Marks:** M1, A1 3
**Guidance:** If consistent "0.7" incorrect of $\frac{1}{7}$, $\frac{2}{3}$ or 0.03 allow M marks in ii, iii & i. M1 in i
### Part ii
**Answer:** $0.7^1$ alone = $\frac{2401}{10000}$ or 0.240 (3 sf)
**Marks:** M1, A1 2
**Guidance:** $1 - (0.3 + 0.7×0.3 + 0.7^2×0.3 + 0.7^3×0.3)$ NB $1 - 0.7^4$ : M0
### Part iii
**Answer:** $1 - 0.7^5$ = 0.832 (3 sfs)
**Marks:** M2, A1 3
**Guidance:** or $0.3 + 0.7×0.3 + ... + 0.7^4×0.3$ M2. M1 for one term extra or omitted or wrong or for $1 - ($ above). M1 for $1 - 0.7^n$ or $0.7^7$. NB Beware: $1 - 0.7^6 = 0.882$
---30% of people own a Talk-2 phone. People are selected at random, one at a time, and asked whether they own a Talk-2 phone. The number of people questioned, up to and including the first person who owns a Talk-2 phone, is denoted by $X$. Find
\begin{enumerate}[label=(\roman*)]
\item P($X = 4$), [3]
\item P($X > 4$), [2]
\item P($X < 6$). [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2009 Q4 [8]}}