| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard two-outcome diagnostic test |
| Difficulty | Moderate -0.3 This is a standard conditional probability question using Bayes' theorem with clearly stated probabilities and a tree diagram scaffold. While it requires careful organization of information and multi-step calculation, it follows a well-established textbook template with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tree diagram with all 6 branches (structure) | M1 | Tree without probabilities or labels |
| \(0.02\) (Disease), \(0.95\) (Positive) on correct branches | A1 | |
| \(0.03\) (Positive) on correct branch | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{Positive Test}) = 0.02 \times 0.95 + 0.98 \times 0.03\) | M1A1ft | M1 for sum of two products, at least one correct from diagram |
| \(= 0.0484\) | A1 | A1ft follows from probabilities on their tree; also accept \(\frac{121}{2500}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{No disease} \mid \text{Positive test}) = \dfrac{0.98 \times 0.03}{0.0484}\) | M1 | M1 for conditional probability with correct numerator and denominator from (b) |
| \(= 0.607438\ldots \approx 0.607\) | A1 | Also accept \(\frac{147}{242}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Test not very useful OR high probability of not having the disease for a person with a positive test | B1 |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tree diagram with all 6 branches (structure) | M1 | Tree without probabilities or labels |
| $0.02$ (Disease), $0.95$ (Positive) on correct branches | A1 | |
| $0.03$ (Positive) on correct branch | A1 | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{Positive Test}) = 0.02 \times 0.95 + 0.98 \times 0.03$ | M1A1ft | M1 for sum of two products, at least one correct from diagram |
| $= 0.0484$ | A1 | A1ft follows from probabilities on their tree; also accept $\frac{121}{2500}$ |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{No disease} \mid \text{Positive test}) = \dfrac{0.98 \times 0.03}{0.0484}$ | M1 | M1 for conditional probability with correct numerator and denominator from (b) |
| $= 0.607438\ldots \approx 0.607$ | A1 | Also accept $\frac{147}{242}$ |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Test not very useful OR high probability of not having the disease for a person with a positive test | B1 | |
---\begin{enumerate}
\item A disease is known to be present in $2 \%$ of a population. A test is developed to help determine whether or not someone has the disease.
\end{enumerate}
Given that a person has the disease, the test is positive with probability 0.95\\
Given that a person does not have the disease, the test is positive with probability 0.03\\
(a) Draw a tree diagram to represent this information.
A person is selected at random from the population and tested for this disease.\\
(b) Find the probability that the test is positive.
A doctor randomly selects a person from the population and tests him for the disease. Given that the test is positive,\\
(c) find the probability that he does not have the disease.\\
(d) Comment on the usefulness of this test.
\hfill \mbox{\textit{Edexcel S1 2008 Q1 [9]}}