| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Standard +0.3 This is a standard S3-level tree diagram question with conditional probability and Bayes' theorem. While it involves multiple branches and several calculations, the structure is straightforward: draw a two-stage tree, multiply along branches, and apply standard formulas. The arithmetic is careful but routine, making it slightly easier than average for A-level. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| First split: Disease (0.1) / No Disease (0.9) | B1 | Correct first branch probabilities |
| For Disease branch: Test A Positive (0.90), Negative (0.02), Inconclusive→Test B (0.08) | B1 | Correct second stage for disease branch |
| For No Disease branch: Test A Positive (0.01), Negative (0.80), Inconclusive→Test B (0.19) | B1 | Correct second stage for no disease branch |
| Test B branches: Disease: Positive (0.98), Negative (0.02); No Disease: Positive (0.01), Negative (0.99) | B1 | Correct Test B probabilities on both branches |
| ## Part (b)(i)(A) - P(tests negative \ | has disease) | |
| Answer | Mark | Guidance |
| \(P(\text{neg} \cap \text{disease}) = 0.1 \times 0.02 + 0.1 \times 0.08 \times 0.02\) | M1 | Adding correct path products |
| \(= 0.002 + 0.00016 = 0.00216\) | A1 | Correct answer |
| ## Part (b)(i)(B) - P(tests positive \ | no disease) | |
| Answer | Mark | Guidance |
| \(P(\text{pos} \cap \text{no disease}) = 0.9 \times 0.01 + 0.9 \times 0.19 \times 0.01\) | M1 | Adding correct path products |
| \(= 0.009 + 0.00171 = 0.01071\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{incorrect}) = 0.00216 + 0.01071 = 0.01287\) | M1 | Sum of both incorrect probabilities |
| \(10000 \times 0.01287 \approx 130\) | A1 | Answer rounded to nearest 10 |
| ## Part (c)(i) - P(disease \ | positive result) | |
| Answer | Mark | Guidance |
| \(P(\text{positive}) = 0.1 \times 0.90 + 0.1 \times 0.08 \times 0.98 + 0.9 \times 0.01 + 0.9 \times 0.19 \times 0.01\) | M1 | Identifying all positive paths |
| \(= 0.09 + 0.00784 + 0.009 + 0.00171 = 0.10855\) | A1 | Correct total probability of positive |
| \(P(\text{disease} \mid \text{positive}) = \dfrac{0.09 + 0.00784}{0.10855}\) | M1 | Correct use of conditional probability formula |
| \(= \dfrac{0.09784}{0.10855} \approx 0.9013\) | A1 | Correct answer |
| ## Part (c)(ii) - P(no disease \ | negative result) | |
| Answer | Mark | Guidance |
| \(P(\text{negative}) = 0.00216 + 0.9 \times 0.80 + 0.9 \times 0.19 \times 0.99\) | M1 | Identifying all negative paths |
| \(= 0.00216 + 0.72 + 0.169\ 02 = 0.891\ 18\) | A1 | Correct total |
| \(P(\text{no disease} \mid \text{negative}) = \dfrac{0.72 + 0.169\ 02}{0.891\ 18}\) | M1 | Correct conditional probability structure |
| \(= \dfrac{0.889\ 02}{0.891\ 18} \approx 0.9976\) | A1 | Correct final answer |
| Fully correct solution with all working shown | A1 | Final accuracy mark |
# Question 4:
## Part (a) - Tree Diagram
| Answer | Mark | Guidance |
|--------|------|----------|
| First split: Disease (0.1) / No Disease (0.9) | B1 | Correct first branch probabilities |
| For Disease branch: Test A Positive (0.90), Negative (0.02), Inconclusive→Test B (0.08) | B1 | Correct second stage for disease branch |
| For No Disease branch: Test A Positive (0.01), Negative (0.80), Inconclusive→Test B (0.19) | B1 | Correct second stage for no disease branch |
| Test B branches: Disease: Positive (0.98), Negative (0.02); No Disease: Positive (0.01), Negative (0.99) | B1 | Correct Test B probabilities on both branches |
## Part (b)(i)(A) - P(tests negative \| has disease)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{neg} \cap \text{disease}) = 0.1 \times 0.02 + 0.1 \times 0.08 \times 0.02$ | M1 | Adding correct path products |
| $= 0.002 + 0.00016 = 0.00216$ | A1 | Correct answer |
## Part (b)(i)(B) - P(tests positive \| no disease)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{pos} \cap \text{no disease}) = 0.9 \times 0.01 + 0.9 \times 0.19 \times 0.01$ | M1 | Adding correct path products |
| $= 0.009 + 0.00171 = 0.01071$ | A1 | Correct answer |
## Part (b)(ii) - Incorrect diagnoses in 10,000
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{incorrect}) = 0.00216 + 0.01071 = 0.01287$ | M1 | Sum of both incorrect probabilities |
| $10000 \times 0.01287 \approx 130$ | A1 | Answer rounded to nearest 10 |
## Part (c)(i) - P(disease \| positive result)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{positive}) = 0.1 \times 0.90 + 0.1 \times 0.08 \times 0.98 + 0.9 \times 0.01 + 0.9 \times 0.19 \times 0.01$ | M1 | Identifying all positive paths |
| $= 0.09 + 0.00784 + 0.009 + 0.00171 = 0.10855$ | A1 | Correct total probability of positive |
| $P(\text{disease} \mid \text{positive}) = \dfrac{0.09 + 0.00784}{0.10855}$ | M1 | Correct use of conditional probability formula |
| $= \dfrac{0.09784}{0.10855} \approx 0.9013$ | A1 | Correct answer |
## Part (c)(ii) - P(no disease \| negative result)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{negative}) = 0.00216 + 0.9 \times 0.80 + 0.9 \times 0.19 \times 0.99$ | M1 | Identifying all negative paths |
| $= 0.00216 + 0.72 + 0.169\ 02 = 0.891\ 18$ | A1 | Correct total |
| $P(\text{no disease} \mid \text{negative}) = \dfrac{0.72 + 0.169\ 02}{0.891\ 18}$ | M1 | Correct conditional probability structure |
| $= \dfrac{0.889\ 02}{0.891\ 18} \approx 0.9976$ | A1 | Correct final answer |
| Fully correct solution with all working shown | A1 | Final accuracy mark |4 It is proposed to introduce, for all males at age 60, screening tests, A and B, for a certain disease.
Test B is administered only when the result of Test A is inconclusive.
It is known that 10\% of 60-year-old men suffer from the disease.
For those 60 -year-old men suffering from the disease:
\begin{itemize}
\item Test A is known to give a positive result, indicating a presence of the disease, in $90 \%$ of cases, a negative result in $2 \%$ of cases and a requirement for the administration of Test B in $8 \%$ of cases;
\item Test B is known to give a positive result in $98 \%$ of cases and a negative result in 2\% of cases.
\end{itemize}
For those 60 -year-old men not suffering from the disease:
\begin{itemize}
\item Test A is known to give a positive result in $1 \%$ of cases, a negative result in $80 \%$ of cases and a requirement for the administration of Test B in 19\% of cases;
\item Test B is known to give a positive result in $1 \%$ of cases and a negative result in $99 \%$ of cases.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to represent the above information.
\item \begin{enumerate}[label=(\roman*)]
\item Hence, or otherwise, determine the probability that:\\
(A) a 60-year-old man, suffering from the disease, tests negative;\\
(B) a 60-year-old man, not suffering from the disease, tests positive.
\item A random sample of ten thousand 60-year-old men is given the screening tests. Calculate, to the nearest 10, the number who you would expect to be given an incorrect diagnosis.
\end{enumerate}\item Determine the probability that:
\begin{enumerate}[label=(\roman*)]
\item a 60-year-old man suffers from the disease given that the tests provide a positive result;
\item a 60-year-old man does not suffer from the disease given that the tests provide a negative result.
\end{itemize}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-09_2484_1709_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-10_2484_1712_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-11_2484_1709_223_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2010 Q4 [13]}}