| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Independence test with conditional probability |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability concepts (independence, conditional probability, Venn diagrams) with standard calculations. Part (i) requires simple comparison of P(T|M) with P(T), part (ii) uses the definition P(T∩M) = P(T|M)×P(M), and part (iii) involves routine arithmetic to fill in a Venn diagram. All steps are direct applications of formulas with no problem-solving insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | Because P(T |
| [1] | Or 0.8 ≠ 0.55 | Or P(T ∩ M) (= 0.264) ≠ P(T) × P(M), |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | P(T ∩ M) = P(T | M) × P(M) = 0.80 × 0.33 = 0.264 |
| Answer | Marks |
|---|---|
| [2] | For product |
| CAO | A0 for 0.26 |
| (iii) | T M |
| Answer | Marks |
|---|---|
| 0.384 | G1 |
| Answer | Marks |
|---|---|
| [3] | For two labelled |
| Answer | Marks |
|---|---|
| and 1 | Allow labels such as P(T) etc |
Question 3:
3 | (i) | Because P(T | M) P(T) | E1
[1] | Or 0.8 ≠ 0.55 | Or P(T ∩ M) (= 0.264) ≠ P(T) × P(M),
provided 0.264 in (ii)
Or 0.264 ≠ 0.55 × 0.33 (=0.1815)
Look out for complement methods, etc
(ii) | P(T ∩ M) = P(T | M) × P(M) = 0.80 × 0.33 = 0.264 | M1
A1
[2] | For product
CAO | A0 for 0.26
(iii) | T M
0.2 86 0.264 0.066
0.384 | G1
G1
G1
[3] | For two labelled
intersecting circles
For at least 2 correct
probabilities. FT their
P(T ∩ M)
For remaining
probabilities. FT their
P(T ∩ M), providing
probabilities between 0
and 1 | Allow labels such as P(T) etc
Allow other shapes in place of circles
No need for ‘box’
FT from 0.1815 in (ii) gives 0.3685,
0.1815, 0.1485, 0.3015In a food survey, a large number of people are asked whether they like tomato soup, mushroom soup, both or neither. One of these people is selected at random.
• $T$ is the event that this person likes tomato soup.
• $M$ is the event that this person likes mushroom soup.
You are given that $\text{P}(T) = 0.55$, $\text{P}(M) = 0.33$ and $\text{P}(T|M) = 0.80$.
\begin{enumerate}[label=(\roman*)]
\item Use this information to show that the events $T$ and $M$ are not independent. [1]
\item Find $\text{P}(T \cap M)$. [2]
\item Draw a Venn diagram showing the events $T$ and $M$, and fill in the probability corresponding to each of the four regions of your diagram. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q3 [6]}}