| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Conditional probability from tree |
| Difficulty | Standard +0.3 This is a standard tree diagram question with straightforward conditional probability calculations. While it has multiple parts and requires careful organization of probabilities, all steps involve direct application of the multiplication and addition rules with no conceptual challenges beyond basic S1 material. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
# Question 1
## (i)
G1 probabilities of result
G1 probabilities of disease
G1 probabilities of clear
G1 labels
**Total: 4 marks**
## (ii)
$P(\text{negative and clear}) = 0.91 \times 0.99 = 0.9009$
M1 for their $0.91 \times 0.99$
A1 CAO
**Total: 2 marks**
## (iii)
$P(\text{has disease}) = 0.03 \times 0.95 + 0.06 \times 0.10 + 0.91 \times 0.01$
$= 0.0285 + 0.006 + 0.0091$
$= 0.0436$
M1 three products
M1 dep sum of three products
A1 FT their tree
**Total: 3 marks**
## (iv)
$P(\text{negative} \mid \text{has disease}) = \dfrac{P(\text{negative and has disease})}{P(\text{has disease})} = \dfrac{0.0091}{0.0436} = 0.2087$
M1 for their $0.01 \times 0.91$ or $0.0091$ on its own or as numerator
M1 indep for their $0.0436$ as denominator
A1 FT their tree
**Total: 3 marks**
## (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' or 'could be improved', etc
E1 FT
**Total: 2 marks**
## (vi)
$P(\text{negative or doubtful and declared clear})$
$= 0.91 + 0.06 \times 0.10 \times 0.02 + 0.06 \times 0.90 \times 1$
$= 0.91 + 0.00012 + 0.054 = 0.96412$
M1 for their $0.91 +$
M1 for either triplet
M1 for second triplet
A1 CAO
**Total: 4 marks**
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**TOTAL: 18 marks**1 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]{f3d936ba-8f60-4350-a5b3-92200996434c-1_833_1156_851_573}\\
(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 Q1 [18]}}