| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard two-outcome diagnostic test |
| Difficulty | Moderate -0.5 This is a standard conditional probability question using tree diagrams and Bayes' theorem. While it has multiple parts and requires careful calculation, all techniques are routine for S1 level—completing a tree diagram, using the law of total probability, and applying Bayes' theorem. The final part involves combining probabilities across independent events, which is straightforward once the earlier parts are understood. Slightly easier than average due to its structured, step-by-step nature with no novel problem-solving required. |
| 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 |
|---|---|---|
| Tree diagram: Genuine (0.9) \(\rightarrow\) Positive (0.95) / Negative (0.05); Fake (0.1) \(\rightarrow\) Positive (0.) / Negative (0.) | G1 for left hand set of branches fully correct including labels and probabilities; G1 for right hand set of branches fully correct | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{test is positive}) = (0.9)(0.95) + (0.1)(0.2) = 0.875\) | M1 two correct pairs added, A1 CAO | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{test is correct}) = (0.9)(0.95) + (0.1)(0.8) = 0.935\) | M1 two correct pairs added, A1 CAO | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Genuine} | \text{Positive}) = \frac{0.855}{0.875} = 0.977\) | M1 numerator; M1 denominator; A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Fake} | \text{Negative}) = \frac{0.08}{0.125} = 0.64\) | M1 numerator; M1 denominator; A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| However, more than a third of those paintings with a negative result are genuine so a further test is needed. | E1 FT; E1 FT | Note: Allow sensible alternative answers; 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0000806\) | M1 for \(0.9 \times 0.05\ (=0.045)\); M1 for complete correct triple product; M1 indep for cubing; A1 CAO | 4 marks |
# Question 4:
## Part (i)
Tree diagram: Genuine (0.9) $\rightarrow$ Positive (0.95) / Negative (0.05); Fake (0.1) $\rightarrow$ Positive (0.) / Negative (0.) | G1 for left hand set of branches fully correct including labels and probabilities; G1 for right hand set of branches fully correct | 2 marks
## Part (ii)
$P(\text{test is positive}) = (0.9)(0.95) + (0.1)(0.2) = 0.875$ | M1 two correct pairs added, A1 CAO | 2 marks
## Part (iii)
$P(\text{test is correct}) = (0.9)(0.95) + (0.1)(0.8) = 0.935$ | M1 two correct pairs added, A1 CAO | 2 marks
## Part (iv)
$P(\text{Genuine}|\text{Positive}) = \frac{0.855}{0.875} = 0.977$ | M1 numerator; M1 denominator; A1 CAO | 3 marks
## Part (v)
$P(\text{Fake}|\text{Negative}) = \frac{0.08}{0.125} = 0.64$ | M1 numerator; M1 denominator; A1 CAO | 3 marks
## Part (vi)
EITHER: A positive test means the painting is almost certain to be genuine so no need for a further test.
However, more than a third of those paintings with a negative result are genuine so a further test is needed. | E1 FT; E1 FT | Note: Allow sensible alternative answers; 2 marks
## Part (vii)
$P(\text{all 3 genuine}) = (0.9 \times 0.05 \times 0.96)^3$
$= (0.045 \times 0.96)^3$
$= (0.0432)^3$
$= 0.0000806$ | M1 for $0.9 \times 0.05\ (=0.045)$; M1 for complete correct triple product; M1 indep for cubing; A1 CAO | 4 marks4 It has been estimated that $90 \%$ of paintings offered for sale at a particular auction house are genuine, and that the other $10 \%$ are fakes. The auction house has a test to determine whether or not a given painting is genuine. If this test gives a positive result, it suggests that the painting is genuine. A negative result suggests that the painting is a fake.
If a painting is genuine, the probability that the test result is positive is 0.95 .\\
If a painting is a fake, the probability that the test result is positive is 0.2 .\\
(i) Copy and complete the probability tree diagram below, to illustrate the information above.\\
\includegraphics[max width=\textwidth, alt={}, center]{f3d936ba-8f60-4350-a5b3-92200996434c-3_466_667_834_737}
Calculate the probabilities of the following events.\\
(ii) The test gives a positive result.\\
(iii) The test gives a correct result.\\
(iv) The painting is genuine, given a positive result.\\
(v) The painting is a fake, given a negative result.
A second test is more accurate, but very expensive. The auction house has a policy of only using this second test on those paintings with a negative result on the original test.\\
(vi) Using your answers to parts (iv) and (v), explain why the auction house has this policy.
The probability that the second test gives a correct result is 0.96 whether the painting is genuine or a fake.\\
(vii) Three paintings are independently offered for sale at the auction house. Calculate the probability that all three paintings are genuine, are judged to be fakes in the first test, but are judged to be genuine in the second test.
\hfill \mbox{\textit{OCR MEI S1 Q4 [18]}}