| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Binomial given fixed number of trials |
| Difficulty | Moderate -0.8 This is a straightforward application of standard probability distributions (binomial, geometric, and negative binomial) with clearly stated parameters (p=0.04). All parts require direct formula application with no problem-solving insight: (a) binomial probability, (b) geometric probability, (c) binomial with r=0, (d) expectation of negative binomial, (e) negative binomial probability. The multi-part structure and context add length but not conceptual difficulty—each part is a textbook exercise testing recall of formulas. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | Using B(50, 0.04) |
| P(X = 2) = 0.276 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 1.1 | BC | |
| 3 | (b) | 0.969×0.04=0.0277 |
| [1] | 1.1 | Allow 0.028 |
| 3 | (c) | 0.9620 =0.442 |
| [1] | 1.1 | |
| 3 | (d) | 1 |
| Answer | Marks |
|---|---|
| = 75 | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.1 |
| Answer | Marks |
|---|---|
| 1.1 | Must quote |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (e) | Require P(2 misunderstood in first 59) × 0.04 |
| Answer | Marks |
|---|---|
| 0.267 × 0.04 = 0.0107 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | For identifying required probability |
Question 3:
3 | (a) | Using B(50, 0.04)
P(X = 2) = 0.276 | M1
A1
[2] | 3.3
1.1 | BC
3 | (b) | 0.969×0.04=0.0277 | B1
[1] | 1.1 | Allow 0.028
3 | (c) | 0.9620 =0.442 | B1
[1] | 1.1
3 | (d) | 1
Expected value for one misunderstood = =25
0.04
Because geometric
For 3 misunderstood expected number = 25 + 25 + 25
= 75 | B1
E1
E1
[3] | 2.1
2.4
1.1 | Must quote
probabilities to get full
marks
3 | (e) | Require P(2 misunderstood in first 59) × 0.04
so using B(59, 0.04) gives P(X = 2) = 0.267
0.267 × 0.04 = 0.0107 | B1
M1
A1
[3] | 3.1a
2.2a
1.1 | For identifying required probability
Use of correct binomial
BC3 In air traffic management, air traffic controllers send radio messages to pilots. On receiving a message, the pilot repeats it back to the controller to check that it has been understood correctly.
At a particular site, on average $4 \%$ of messages sent by controllers are not repeated back correctly and so have been misunderstood. You should assume that instances of messages being misunderstood occur randomly and independently.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that exactly 2 messages are misunderstood in a sequence of 50 messages.
\item Find the probability that in a sequence of messages, the 10th message is the first one which is misunderstood.
\item Find the probability that in a sequence of 20 messages, there are no misunderstood messages.
\item Determine the expected number of messages required for 3 of them to be misunderstood.
\item Determine the probability that in a sequence of messages, the 3rd misunderstood message is the 60th message in the sequence.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2021 Q3 [10]}}