| Exam Board | WJEC |
|---|---|
| Module | Unit 4 (Unit 4) |
| Session | Specimen |
| Marks | 6 |
| 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 problem using Bayes' theorem with clearly stated probabilities. Part (a) is straightforward calculation, part (b) requires the law of total probability, and part (c) applies Bayes' theorem. While it requires careful organization (tree diagram helps), it's a textbook application with no novel insight needed—slightly easier than average due to its routine nature. |
| 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 |
|---|---|---|
| \(\text{Prob} = 0.96 \times 0.99 = 0.9504\) | M1, A1 | AO2, AO1 |
| \(P(B) = 0.04 \times 0.98 + 0.96 \times 0.01 = 0.0488\) | M1, A1 | AO3, AO1 |
| \(P(A\ | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.04 \times 0.98}{0.0488} = 0.803(278688\ldots)\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Prob} = 0.96 \times 0.99 = 0.9504\) | (M1), (A1) | (AO1), (AO2) |
| $\text{Prob} = 0.96 \times 0.99 = 0.9504$ | M1, A1 | AO2, AO1 | diagram |
| $P(B) = 0.04 \times 0.98 + 0.96 \times 0.01 = 0.0488$ | M1, A1 | AO3, AO1 | |
| $P(A\|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.04 \times 0.98}{0.0488} = 0.803(278688\ldots)$ | M1, A1 | AO3, AO1 | |
**Alternative mark scheme for (a):**
| $\text{Prob} = 0.96 \times 0.99 = 0.9504$ | (M1), (A1) | (AO1), (AO2) | |\begin{enumerate}
\item It is known that $4 \%$ of a population suffer from a certain disease. When a diagnostic test is applied to a person with the disease, it gives a positive response with probability 0.98 . When the test is applied to a person who does not have the disease, it gives a positive response with probability 0.01 .\\
(a) Using a tree diagram, or otherwise, show that the probability of a person who does not have the disease giving a negative response is 0.9504 .
\end{enumerate}
The test is applied to a randomly selected member of the population.\\
(b) Find the probability that a positive response is obtained.\\
(c) Given that a positive response is obtained, find the probability that the person has the disease.\\
\hfill \mbox{\textit{WJEC Unit 4 Q1 [6]}}