| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question using the law of total probability and Bayes' theorem with clearly presented data. Part (a) involves routine calculations with given probabilities, while part (b) requires understanding of conditional probability with multiple events but follows standard S3 techniques. The structure is typical of S3 exam questions with no novel insights required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Time | ||||
| \(\boldsymbol { \leq } \mathbf { 1 }\) hour | \(\leq \mathbf { 2 4 }\) hours | \(\leq 72\) hours | ||
| Station | Alpha | 55 | 80 | 100 |
| Beta | 60 | 85 | 100 | |
| Gamma | 40 | 75 | 100 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{resolved at Gamma}) = 0.15 \times 1.00 = 0.15\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{resolved at Alpha within 1 hour}) = 0.60 \times 0.55 = 0.33\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{resolved within 24 hours}) = 0.60 \times 0.80 + 0.25 \times 0.85 + 0.15 \times 0.75\) | M1 | Correct structure |
| \(= 0.48 + 0.2125 + 0.1125 = 0.805\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{Beta} \mid \text{resolved within 24 hours}) = \frac{P(\text{Beta and within 24 hours})}{P(\text{within 24 hours})}\) | M1 | Conditional probability structure |
| \(= \frac{0.25 \times 0.85}{0.805}\) | A1 | Correct numerator |
| \(= \frac{0.2125}{0.805} = 0.2640\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{all 3 at Beta} \mid \text{all resolved within 24h}) = \left(\frac{0.2125}{0.805}\right)^3\) | M1 | Correct method using conditional probability |
| \(= (0.2640)^3 = 0.01842\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{all same station} \mid \text{all within 24h})\) | M1 | Sum of three terms |
| \(= \left(\frac{0.48}{0.805}\right)^3 + \left(\frac{0.2125}{0.805}\right)^3 + \left(\frac{0.1125}{0.805}\right)^3\) | A1 A1 | Each probability correct |
| \(= (0.5963)^3 + (0.2640)^3 + (0.1398)^3\) | M1 | Correct cubing and adding |
| \(= 0.2121 + 0.01842 + 0.002732 = 0.2332\) | A1 | Correct final answer |
# Question 3:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{resolved at Gamma}) = 0.15 \times 1.00 = 0.15$ | B1 | |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{resolved at Alpha within 1 hour}) = 0.60 \times 0.55 = 0.33$ | B1 | |
## Part (a)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{resolved within 24 hours}) = 0.60 \times 0.80 + 0.25 \times 0.85 + 0.15 \times 0.75$ | M1 | Correct structure |
| $= 0.48 + 0.2125 + 0.1125 = 0.805$ | A1 | Correct answer |
## Part (a)(iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{Beta} \mid \text{resolved within 24 hours}) = \frac{P(\text{Beta and within 24 hours})}{P(\text{within 24 hours})}$ | M1 | Conditional probability structure |
| $= \frac{0.25 \times 0.85}{0.805}$ | A1 | Correct numerator |
| $= \frac{0.2125}{0.805} = 0.2640$ | A1 | Correct final answer |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{all 3 at Beta} \mid \text{all resolved within 24h}) = \left(\frac{0.2125}{0.805}\right)^3$ | M1 | Correct method using conditional probability |
| $= (0.2640)^3 = 0.01842$ | A1 | Correct answer |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{all same station} \mid \text{all within 24h})$ | M1 | Sum of three terms |
| $= \left(\frac{0.48}{0.805}\right)^3 + \left(\frac{0.2125}{0.805}\right)^3 + \left(\frac{0.1125}{0.805}\right)^3$ | A1 A1 | Each probability correct |
| $= (0.5963)^3 + (0.2640)^3 + (0.1398)^3$ | M1 | Correct cubing and adding |
| $= 0.2121 + 0.01842 + 0.002732 = 0.2332$ | A1 | Correct final answer |3 An IT help desk has three telephone stations: Alpha, Beta and Gamma. Each of these stations deals only with telephone enquiries.
The probability that an enquiry is received at Alpha is 0.60 .\\
The probability that an enquiry is received at Beta is 0.25 .\\
The probability that an enquiry is received at Gamma is 0.15 .
Each enquiry is resolved at the station that receives the enquiry. The percentages of enquiries resolved within various times at each station are shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{} & \multicolumn{3}{|c|}{Time} \\
\hline
& & $\boldsymbol { \leq } \mathbf { 1 }$ hour & $\leq \mathbf { 2 4 }$ hours & $\leq 72$ hours \\
\hline
Station & Alpha & 55 & 80 & 100 \\
\hline
& Beta & 60 & 85 & 100 \\
\hline
& Gamma & 40 & 75 & 100 \\
\hline
\end{tabular}
\end{center}
For example:\\
80 per cent of enquiries received at Alpha are resolved within 24 hours;\\
25 per cent of enquiries received at Alpha take between 1 hour and 24 hours to resolve.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that an enquiry, selected at random, is:
\begin{enumerate}[label=(\roman*)]
\item resolved at Gamma;
\item resolved at Alpha within 1 hour;
\item resolved within 24 hours;
\item received at Beta, given that it is resolved within 24 hours.
\end{enumerate}\item A random sample of 3 enquiries was selected.
Given that all 3 enquiries were resolved within 24 hours, calculate the probability that they were all received at:
\begin{enumerate}[label=(\roman*)]
\item Beta;
\item the same station.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
& & \\
\hline
\end{tabular}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q3 [13]}}