| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Three-outcome diagnostic test |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question using tree diagrams with clearly stated probabilities. Parts (i)-(v) involve basic probability calculations (multiplication and addition along branches), while part (vi) requires combining two mutually exclusive events. The structure is standard for S1 with no conceptual surprises, though the multi-part nature and careful reading required elevate it slightly above the most routine questions. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks |
|---|---|
| G1 probabilities of result, G1 probabilities of disease, G1 probabilities of clear, G1 labels | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{negative and clear}) = 0.91\times0.99 = 0.9009\) | M1 for their \(0.91\times0.99\), A1 CAO | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0285 + 0.006 + 0.0091 = 0.0436\) | M1 three products, M1dep sum of three products, A1 FT their tree | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{negative} \mid \text{has disease}) = \frac{P(\text{negative and has disease})}{P(\text{has disease})} = \frac{0.0091}{0.0436} = 0.2087\) | M1 for their \(0.01\times0.91\) or \(0.0091\) as numerator, M1dep for their \(0.0436\) as denominator, A1 FT their tree | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Thus the test result is not very reliable. A relatively large proportion of people who have the disease will test negative. | E1 FT for idea of 'not reliable', E1 FT | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.91 + 0.00012 + 0.054 = 0.96412\) | M1 for their \(0.91\) + M1 for either triplet, M1 for second triplet, A1 CAO | 4 marks |
# Question 7:
## Part (i)
Tree diagram with:
- Branch probabilities: $0.03$ (positive result), $0.06$ (doubtful result), $0.91$ (negative result)
- For positive: $0.95$ (has disease), $0.05$ (clear)
- For doubtful: $0.10$ (has disease), $0.90$ (clear)
- For negative: $0.01$ (has disease), $0.99$ (clear)
| G1 probabilities of result, G1 probabilities of disease, G1 probabilities of clear, G1 labels | 4 marks
## Part (ii)
$P(\text{negative and clear}) = 0.91\times0.99 = 0.9009$ | M1 for their $0.91\times0.99$, A1 CAO | 2 marks
## Part (iii)
$P(\text{has disease}) = 0.03\times0.95 + 0.06\times0.10 + 0.91\times0.01$
$= 0.0285 + 0.006 + 0.0091 = 0.0436$ | M1 three products, M1dep sum of three products, A1 FT their tree | 3 marks
## Part (iv)
$P(\text{negative} \mid \text{has disease}) = \frac{P(\text{negative and has disease})}{P(\text{has disease})} = \frac{0.0091}{0.0436} = 0.2087$ | M1 for their $0.01\times0.91$ or $0.0091$ as numerator, M1dep for their $0.0436$ as denominator, A1 FT their tree | 3 marks
## Part (v)
Thus the test result is not very reliable. A relatively large proportion of people who have the disease will test negative. | E1 FT for idea of 'not reliable', E1 FT | 2 marks
## Part (vi)
$P(\text{negative or doubtful and declared clear})$
$= 0.91 + 0.06\times0.10\times0.02 + 0.06\times0.90\times1$
$= 0.91 + 0.00012 + 0.054 = 0.96412$ | M1 for their $0.91$ + M1 for either triplet, M1 for second triplet, A1 CAO | 4 marks
---7 A screening test for a particular disease is applied to everyone in a large population. The test classifies people into three groups: 'positive', 'doubtful' and 'negative'. Of the population, $3 \%$ is classified as positive, $6 \%$ as doubtful and the rest negative.
In fact, of the people who test positive, only $95 \%$ have the disease. Of the people who test doubtful, $10 \%$ have the disease. Of the people who test negative, $1 \%$ actually have the disease.
People who do not have the disease are described as 'clear'.\\
(i) Copy and complete the tree diagram to show this information.\\
\includegraphics[max width=\textwidth, alt={}, center]{5e4f3310-b96e-43db-9b6d-61da3270db06-5_830_1157_845_536}\\
(ii) Find the probability that a randomly selected person tests negative and is clear.\\
(iii) Find the probability that a randomly selected person has the disease.\\
(iv) Find the probability that a randomly selected person tests negative given that the person has the disease.\\
(v) Comment briefly on what your answer to part (iv) indicates about the effectiveness of the screening test.
Once the test has been carried out, those people who test doubtful are given a detailed medical examination. If a person has the disease the examination will correctly identify this in $98 \%$ of cases. If a person is clear, the examination will always correctly identify this.\\
(vi) A person is selected at random. Find the probability that this person either tests negative originally or tests doubtful and is then cleared in the detailed medical examination.
\hfill \mbox{\textit{OCR MEI S1 2007 Q7 [18]}}