| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Find minimum n for P(X ≤ n) > threshold |
| Difficulty | Standard +0.3 Part (i) requires setting up and solving (1-0.88)^n > 0.95 using logarithms, which is a standard S1 technique. Part (ii) involves recognizing and applying the negative binomial distribution formula, which is more conceptually demanding but still a bookwork application. Both parts are routine for S1 students who know the distributions, making this slightly easier than average overall. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
### Part i
**Answer:** $(1 - 0.12)^n$. log 0.05. log 0.88
**Marks:** M1, M1, A1 3
**Guidance:** or $0.88^{23} = 0.052...$ or $0.88^{24} = 0.046...$. Can be implied by 2nd M1 allow $n - 1$. or $log_{0.88}0.05$ or 23.4(...). Ignore incorrect inequ or equals signs
$n = 24$
### Part ii
**Answer:** $^nC_2 × 0.88^n × 0.12^2$ (= 0.1295...)
**Marks:** M3, M2
**Guidance:** or $0.88^n × 0.12^2$ or $^nC_2 × 0.88^n × 0.12^2$ + extra or 2 successes in 6 trials implied or $^nC_2$ (M1). dep $\geq$ M1 (A1 5). $0.88^n×0.12^2×0.12$: M2M1. $0.88^n × 0.12^3$: M0M0A0. unless clear P(2 success in 6 trials) $× 0.12$ in which case M2M1A0
$× 0.12 = 0.0155$
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## Total Marks: 72Repeated independent trials of a certain experiment are carried out. On each trial the probability of success is 0.12.
\begin{enumerate}[label=(\roman*)]
\item Find the smallest value of $n$ such that the probability of at least one success in $n$ trials is more than 0.95. [3]
\item Find the probability that the 3rd success occurs on the 7th trial. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2009 Q9 [8]}}