| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric distribution formulas with p=0.08. Part (i) requires direct substitution into P(X=k)=(1-p)^(k-1)×p and simple addition. Part (ii) uses complement rule P(at least 1)=1-P(none)=1-(0.92)^20. All calculations are routine with no conceptual challenges or problem-solving required beyond recognizing the standard geometric distribution setup. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{third selected}) = 0.92^2 \times 0.08 = 0.0677\) or \(= \frac{1058}{15625}\) | M1 | For \(0.92^2\); with \(p+q=1\); with no extra terms |
| M1 | For \(p^2 \times q\) | |
| A1 [3] | CAO; allow 0.068 but not 0.067 nor 0.07; SC1 for 'without replacement' method \(\frac{92}{100}\times\frac{91}{99}\times\frac{8}{98} = 0.0690\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{second}) + P(\text{third}) = (0.92 \times 0.08) + (0.92^2 \times 0.08)\) | M1 | For \(0.92 \times 0.08\); with no extra terms |
| \(= 0.0736 + 0.0677 = 0.1413 = \frac{2208}{15625}\) | A1 [2] | FT their 0.0677; allow 0.141 to 0.142 and allow 0.14 with working; SC1 for answer of 0.143 from 'without replacement' method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{at least one of first 20}) = 1 - P(\text{none of first 20})\) | M1 | \(0.92^{20}\); accept answer of 0.81 or better from \(P(1)+P(2)+\ldots\), or SC2 if all correct working shown but wrong answer; no marks for 'without replacement' method |
| \(= 1 - 0.92^{20} = 1 - 0.1887 = 0.8113\) | M1 | \(1 - 0.92^{20}\) |
| A1 [3] | CAO; allow 0.81 with working but not 0.812 |
# Question 4(i)(A):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{third selected}) = 0.92^2 \times 0.08 = 0.0677$ or $= \frac{1058}{15625}$ | M1 | For $0.92^2$; with $p+q=1$; with no extra terms |
| | M1 | For $p^2 \times q$ |
| | A1 [3] | CAO; allow 0.068 but not 0.067 nor 0.07; SC1 for 'without replacement' method $\frac{92}{100}\times\frac{91}{99}\times\frac{8}{98} = 0.0690$ |
# Question 4(i)(B):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{second}) + P(\text{third}) = (0.92 \times 0.08) + (0.92^2 \times 0.08)$ | M1 | For $0.92 \times 0.08$; with no extra terms |
| $= 0.0736 + 0.0677 = 0.1413 = \frac{2208}{15625}$ | A1 [2] | FT their 0.0677; allow 0.141 to 0.142 and allow 0.14 with working; SC1 for answer of 0.143 from 'without replacement' method |
# Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{at least one of first 20}) = 1 - P(\text{none of first 20})$ | M1 | $0.92^{20}$; accept answer of 0.81 or better from $P(1)+P(2)+\ldots$, or SC2 if all correct working shown but wrong answer; no marks for 'without replacement' method |
| $= 1 - 0.92^{20} = 1 - 0.1887 = 0.8113$ | M1 | $1 - 0.92^{20}$ |
| | A1 [3] | CAO; allow 0.81 with working but not 0.812 |4 It is known that $8 \%$ of the population of a large city use a particular web browser. A researcher wishes to interview some people from the city who use this browser. He selects people at random, one at a time.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the first person that he finds who uses this browser is\\
(A) the third person selected,\\
(B) the second or third person selected.
\item Find the probability that at least one of the first 20 people selected uses this browser.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2012 Q4 [8]}}