| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Multi-type disease diagnostic |
| Difficulty | Challenging +1.2 This is a multi-step conditional probability problem requiring tree diagram completion, simultaneous equations from total probability, and Bayes' theorem. While it involves several calculations and careful bookkeeping across multiple disease types, the techniques are standard S1 material (tree diagrams, law of total probability, conditional probability). The structure is clearly guided by parts (a)-(c), and students who understand the basic formulas can work through systematically. It's moderately harder than average due to the algebraic manipulation required in part (b) and the need to condition on new information in part (c), but doesn't require novel insight beyond applying standard methods. |
| Spec | 2.03a Mutually exclusive and independent events2.03b 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 |
| 0.02 and 0.98 – p correctly placed [no mixing of % and probability] 0.96 and 0.05 plus 1 – q, 0.04, 0.95 correctly placed | B1, B1 | |
| \(\text{P}(T') = pq + 0.02 \times 0.96 + (0.98 - p) \times 0.05 = 0.169\) | M1; A1 | For attempt to form eq'n in p and q using P(T') = 0.169 [at least 2 of 3 correct prod's]. For a fully correct equation in p and q or possibly just q (using their p see 3rd M1) |
| \([ pq - 0.05p = 0.1008]\) | ||
| \(\text{P(do not have disease | T')} = \frac{(0.98 - p) \times 0.05}{0.169} = \frac{41}{169}\) | M1A1ft, A1 |
| So \(p = \mathbf{0.16}\) | dM1, A1 | For substituting their p into an equation for q (ft their p value). For q = 0.68 (or exact equivalent) |
| Should find test useful, doctor knows there is a much greater chance that the person has type A (0.73 compared to 0.16 or 0.163…[from \(\frac{113}{689}\)]) | B1 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| 0.02 and 0.98 – p correctly placed [no mixing of % and probability] 0.96 and 0.05 plus 1 – q, 0.04, 0.95 correctly placed | B1, B1 | |
| $\text{P}(T') = pq + 0.02 \times 0.96 + (0.98 - p) \times 0.05 = 0.169$ | M1; A1 | For attempt to form eq'n in p and q using P(T') = 0.169 [at least 2 of 3 correct prod's]. For a fully correct equation in p and q or possibly just q (using their p see 3rd M1) |
| $[ pq - 0.05p = 0.1008]$ | | |
| $\text{P(do not have disease | T')} = \frac{(0.98 - p) \times 0.05}{0.169} = \frac{41}{169}$ | M1A1ft, A1 | For use of a conditional prob (ratio of probabilities with num or den correct, allow it on num) **and** $\frac{41}{169}$ to form an equation in p. For a correct equation using values from their tree diagram. For awrt 0.726 (or exact fraction $\frac{41}{169}$) |
| So $p = \mathbf{0.16}$ | dM1, A1 | For substituting their p into an equation for q (ft their p value). For q = **0.68** (or exact equivalent) |
| | | |
| Should find test useful, doctor knows there is a much greater chance that the person has type A (0.73 compared to 0.16 or 0.163…[from $\frac{113}{689}$]) | B1 | |
**Notes:**
**(a)** 1st B1 for remainder of 1st column probabilities (allow use of correct p so 0.82)
2nd B1 for remainder of 2nd column probabilities (allow use of correct q so 0.68 and 0.32)
**In (b) or (c) if p or q are used as ft in M or A marks they must be probabilities**
**(b)** 1st M1 for attempt to form eq'n in p and q using P(T') = 0.169 [at least 2 of 3 correct prod's]
1st A1 for a fully correct equation in p and q or possibly just q (using their p see 3rd M1)
2nd M1 for use of a conditional prob (ratio of probabilities with num or den correct, allow it on num) **and** $\frac{41}{169}$ to form an equation in p
2nd A1ft for a correct equation using values from their tree diagram
3rd A1 for solving to get p = 0.16 (or exact equivalent)
3rd dM1 (dep on 1st M1) for substituting their p into an equation for q (ft their p value)
4th A1 for q = 0.68 (or exact equivalent)
**(c)(i)** M1 for an attempt at a conditional prob with numerator of their pq (num < denom)
1st A1ft for a correct ratio of probs (ft their values for p or q with at least one correct)
2nd A1 for awrt 0.726 (or exact fraction $\frac{41}{169}$)
**(c)(ii)** B1 If (c)(i) < 0.7 then B0 for suggesting test should be useful (accept "yes") plus statement: about increased prob or "more likely to have type A than no disease" or "prob of A is high"
---3. A certain disease occurs in a population in 2 mutually exclusive types.
It is difficult to diagnose people with type $A$ of the disease and there is an unknown proportion $p$ of the population with type $A$.\\
It is easier to diagnose people with type $B$ of the disease and it is known that $2 \%$ of the population have type $B$.
A test has been developed to help diagnose whether or not a person has the disease. The event $T$ represents a positive result on the test. After a large-scale trial of the test, the following information was obtained.
For a person with type $B$ of the disease the probability of a positive test result is 0.96 For a person who does not have the disease the probability of a positive test result is 0.05 For a person with type $A$ of the disease the probability of a positive test result is $q$
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac7647f-b291-4a64-9518-fa6438a0cc7d-08_776_965_1050_484}
The probability of a randomly selected person having a positive test result is 0.169 For a person with a positive test result, the probability that they do not have the disease is $\frac { 41 } { 169 }$
\item Find the value of $p$ and the value of $q$.
A doctor is about to see a person who she knows does not have type $B$ of the disease but does have a positive test result.
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that this person has type $A$ of the disease.
\item State, giving a reason, whether or not the doctor will find the test useful.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2019 Q3 [13]}}