| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Mean/expectation of geometric distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of standard geometric distribution formulas with p=1/20. Part (i) requires direct substitution into P(X=6) and P(X>10), while part (ii) asks for the mean E(X)=1/p. All formulas are given in the S1 formula booklet, making this easier than average with minimal problem-solving required. |
| 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 |
5 On average 1 in 20 members of the population of this country has a particular DNA feature. Members of the population are selected at random until one is found who has this feature.\\
(i) Find the probability that the first person to have this feature is
\begin{enumerate}[label=(\alph*)]
\item the sixth person selected,
\item not among the first 10 people selected.\\
(ii) Find the expected number of people selected.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2005 Q5 [8]}}