| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Moderate -0.3 This is a standard conditional probability question using the law of total probability and Bayes' theorem with clearly organized information. Parts (a)(i) and (a)(ii) are straightforward applications of multiplication and addition rules, while (a)(iii) and (b) require Bayes' theorem but with all necessary probabilities already computed. The structure is typical S3 material with no conceptual surprises, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(G \cap I) = 0.5 \times 0.9 = 0.45\) | B1 | CAO; or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(I) = \text{(i)} + P(E \cap I) + P(F \cap I)\) | M1 | 3 possibilities |
| \(= 0.45 + (0.2 \times 0.6) + (0.3 \times 0.75)\) | A1 | \(\geq 1\) correct new term |
| \(= 0.45 + 0.12 + 0.225 = 0.795\) | A1 | CAO; or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(G \mid I) = \frac{P(G \cap I)}{P(I)}\) | M1 | Attempted use of Bayes' Theorem |
| \(= \frac{\text{(i)}}{\text{(ii)}} = \frac{0.45}{0.795} = 0.566\) | m1, A1 | AWRT; or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(E \mid SD) = \frac{P(E \cap SD)}{P(SD)}\) | M1 | Correct use of Bayes' Theorem |
| \(= \frac{0.2 \times 0.25}{(0.2 \times 0.25)+(0.3 \times 0.15)}\) | A1 | Numerator (B1 if no Bayes' Theorem) |
| \(= \frac{0.05}{0.05 + 0.045}\) | A1 | Denominator (B1 if no Bayes' Theorem) |
| \(= \frac{0.05}{0.095} = 0.526\) | A1 | AWRT; or equivalent |
# Question 3:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(G \cap I) = 0.5 \times 0.9 = 0.45$ | B1 | CAO; or equivalent |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(I) = \text{(i)} + P(E \cap I) + P(F \cap I)$ | M1 | 3 possibilities |
| $= 0.45 + (0.2 \times 0.6) + (0.3 \times 0.75)$ | A1 | $\geq 1$ correct **new** term |
| $= 0.45 + 0.12 + 0.225 = 0.795$ | A1 | CAO; or equivalent |
## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(G \mid I) = \frac{P(G \cap I)}{P(I)}$ | M1 | Attempted use of Bayes' Theorem |
| $= \frac{\text{(i)}}{\text{(ii)}} = \frac{0.45}{0.795} = 0.566$ | m1, A1 | AWRT; or equivalent |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(E \mid SD) = \frac{P(E \cap SD)}{P(SD)}$ | M1 | Correct use of Bayes' Theorem |
| $= \frac{0.2 \times 0.25}{(0.2 \times 0.25)+(0.3 \times 0.15)}$ | A1 | Numerator (B1 if no Bayes' Theorem) |
| $= \frac{0.05}{0.05 + 0.045}$ | A1 | Denominator (B1 if no Bayes' Theorem) |
| $= \frac{0.05}{0.095} = 0.526$ | A1 | AWRT; or equivalent |
---3 Each enquiry received by a business support unit is dealt with by Ewan, Fay or Gaby. The probabilities of them dealing with an enquiry are $0.2,0.3$ and 0.5 respectively.
Of enquiries dealt with by Ewan, 60\% are answered immediately, 25\% are answered later the same day and the remainder are answered at a later date.
Of enquiries dealt with by Fay, 75\% are answered immediately, 15\% are answered later the same day and the remainder are answered at a later date.
Of enquiries dealt with by Gaby, 90\% are answered immediately and the remainder are answered at a later date.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that an enquiry:
\begin{enumerate}[label=(\roman*)]
\item is dealt with by Gaby and answered immediately;
\item is answered immediately;
\item is dealt with by Gaby, given that it is answered immediately.
\end{enumerate}\item Determine the probability that an enquiry is dealt with by Ewan, given that it is answered later the same day.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2006 Q3 [11]}}