| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard two-outcome diagnostic test |
| Difficulty | Standard +0.3 This is a straightforward application of Bayes' theorem with clearly stated probabilities. Part (i) requires a standard tree diagram or formula calculation, while part (ii) involves simple algebraic rearrangement. The setup is explicit with no hidden complexity, making it slightly easier than average despite being a multi-part question. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.32 \times 0.96\) or \(0.68 \times 0.08\) | M1 | Allow M marks for 0.8 instead of 0.08 or incorrect 1-0.68 |
| Both, added | M1 | |
| \(= 0.3616\) | A1 | May be implied |
| \(0.32 \times 0.96 \div \text{"0.3616"}\) | M1 | |
| \(0.850\) | A1 | Allow \(0.85\) or \(\frac{96}{113}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{0.32 \times 0.96}{0.32 \times 0.96 + 0.68 \times p} = 0.95\) | M1, A1 | Allow 0.3072 |
| Solve | M1 | Allow failure to multiply brackets correctly, but NOT divide instead of subtract or vice versa |
| \(p = 0.0238\), so \(m = 2.38\) | A1 | \(\frac{192}{1075}\) |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.32 \times 0.96$ or $0.68 \times 0.08$ | M1 | Allow M marks for 0.8 instead of 0.08 or incorrect 1-0.68 |
| Both, added | M1 | |
| $= 0.3616$ | A1 | May be implied |
| $0.32 \times 0.96 \div \text{"0.3616"}$ | M1 | |
| $0.850$ | A1 | Allow $0.85$ or $\frac{96}{113}$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{0.32 \times 0.96}{0.32 \times 0.96 + 0.68 \times p} = 0.95$ | M1, A1 | Allow 0.3072 |
| Solve | M1 | Allow failure to multiply brackets correctly, but NOT divide instead of subtract or vice versa |
| $p = 0.0238$, so $m = 2.38$ | A1 | $\frac{192}{1075}$ |
---2 During an outbreak of a disease, it is known that $68 \%$ of people do not have the disease. Of people with the disease, $96 \%$ react positively to a test for diagnosing it, as do $m \%$ of people who do not have the disease.\\
(i) In the case $m = 8$, find the probability that a randomly chosen person has the disease, given that the person reacts positively to the test.\\
(ii) What value of $m$ would be required for the answer to part (i) to be 0.95 ?
\hfill \mbox{\textit{OCR S4 2014 Q2 [9]}}