| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Moderate -0.8 This is a straightforward probability question testing basic concepts (Venn diagrams, independence, conditional probability) with standard calculations. All parts follow directly from given information using formulae that students are expected to know. No problem-solving insight or multi-step reasoning required—purely routine application of S1 content. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks |
|---|---|
| (i) | W F |
| Answer | Marks |
|---|---|
| 0.56 | G1 for two labelled |
| Answer | Marks | Guidance |
|---|---|---|
| correct probabilities | 3 | Allow labels such as P(W) and P(F) |
| Answer | Marks |
|---|---|
| (ii) | P(W) P(F) = 0.14 0.41 = 0.0574 P(W∩F) = 0.11 |
| So not independent. | M1 for 0.41 0.14 |
| Answer | Marks | Guidance |
|---|---|---|
| P(W∩F) or 0.11 | 2 | Answer of 0.574 gets Max M1A0 |
| Answer | Marks |
|---|---|
| P(W | F) = 0.11/0.41 = 0.268 ≠ P(W) (= 0.14) M1 for full |
| Answer | Marks |
|---|---|
| P(F | W) = 0.11/0.14 = 0.786 ≠ P(F) (= 0.41) M1 for full |
| Answer | Marks |
|---|---|
| (iii) | P(W F) 0.11 11 |
| P(W | F) 0.268 |
| Answer | Marks |
|---|---|
| works (part time), given that the respondent is female. | M1 for correct fraction |
| Answer | Marks | Guidance |
|---|---|---|
| and W | 3 | Allow 0.27 with working |
| Answer | Marks |
|---|---|
| TOTAL | 8 |
Question 5:
--- 5
(i) ---
5
(i) | W F
.03 0.11 0.30
0.56 | G1 for two labelled
intersecting circles
G1 for at least 2 correct
probabilities.
G1 for remaining
correct probabilities | 3 | Allow labels such as P(W) and P(F)
Allow other sensible shapes in place of circles
(ii) | P(W) P(F) = 0.14 0.41 = 0.0574 P(W∩F) = 0.11
So not independent. | M1 for 0.41 0.14
A1 Condone dependent
Must have full method
www
Must have either
P(W∩F) or 0.11 | 2 | Answer of 0.574 gets Max M1A0
Omission of 0.0574 gets M1A0 Max
Or:
P(W|F) = 0.11/0.41 = 0.268 ≠ P(W) (= 0.14) M1 for full
working
P(F|W) = 0.11/0.14 = 0.786 ≠ P(F) (= 0.41) M1 for full
working
No marks without correct working
(iii) | P(W F) 0.11 11
P(W |F) 0.268
P(F) 0.41 41
This is the probability that a randomly selected respondent
works (part time), given that the respondent is female. | M1 for correct fraction
A1
E1
For E1 must be in
context – not just
talking about events F
and W | 3 | Allow 0.27 with working
Allow 11/41 as final answer
Condone ‘if’ or ‘when’ for ‘given that’ but not the words
‘and’ or ‘because’ or ‘due to’ for E1.
E1 (independent of M1): the order/structure must be
correct i.e. no reverse statement
Allow ‘The probability that a randomly selected
female respondent works part time’ oe
TOTAL | 8In a recent survey, a large number of working people were asked whether they worked full-time or part-time, with part-time being defined as less than 25 hours per week. One of the respondents is selected at random.
• $W$ is the event that this person works part-time.
• $F$ is the event that this person is female.
You are given that $\text{P}(W) = 0.14$, $\text{P}(F) = 0.41$ and $\text{P}(W \cap F) = 0.11$.
\begin{enumerate}[label=(\roman*)]
\item Draw a Venn diagram showing the events $W$ and $F$, and fill in the probability corresponding to each of the four regions of your diagram. [3]
\item Determine whether the events $W$ and $F$ are independent. [2]
\item Find $\text{P}(W|F)$ and explain what this probability represents. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q5 [8]}}