| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Geometric with multiple success milestones |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric and binomial probability with clearly defined scenarios. Part (i) requires basic geometric distribution calculations (0.92² × 0.08 and similar), while part (ii) uses complement rule with binomial probability (1 - 0.92²⁰). All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and need for careful probability reasoning. |
| Spec | 2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | (A) |
| Or = 1058/15625 | M1 |
| Answer | Marks |
|---|---|
| [3] | For 0.922 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | (B) | P (second) + P(third) |
| Answer | Marks |
|---|---|
| = 2208/15625 | M1 |
| Answer | Marks |
|---|---|
| [2] | For 0.92 × 0.08 With no extra terms |
| Answer | Marks |
|---|---|
| (ii) | P(At least one of first 20) = 1 - P(None of first 20) |
| = 1 – 0.9220 = 1 – 0.1887 = 0.8113 | M1 |
| Answer | Marks |
|---|---|
| [3] | 0.9220 |
| Answer | Marks |
|---|---|
| CAO | Accept answer of 0.81 or better |
Question 1:
1 | (i) | (A) | P(third selected) = 0.922 × 0.08 = 0.0677
Or = 1058/15625 | M1
M1
A1
[3] | For 0.922
For p2 × q With p + q = 1
With no extra terms
CAO Allow 0.068 but not 0.067 nor
0.07
SC1 for ‘without replacement’ method 92/100×91/99×8/98
=0.0690
(i) | (B) | P (second) + P(third)
= (0.92 × 0.08) + (0.922 × 0.08)
= 0.0736 + 0.0677 = 0.1413
= 2208/15625 | M1
A1
[2] | For 0.92 × 0.08 With no extra terms
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
(ii) | P(At least one of first 20) = 1 - P(None of first 20)
= 1 – 0.9220 = 1 – 0.1887 = 0.8113 | M1
M1
A1
[3] | 0.9220
1 – 0.9220
CAO | Accept answer of 0.81 or better
from P(1) + P(2) + ... , or SC2 if
all correct working shown but
wrong answer
No marks for ‘without
replacement’ method’
Allow 0.81 with working but
not 0.812It 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
\begin{enumerate}[label=(\Alph*)]
\item the third person selected, [3]
\item the second or third person selected. [2]
\end{enumerate}
\item Find the probability that at least one of the first 20 people selected uses this browser. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q1 [8]}}