| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.5 This is a straightforward tree diagram question with standard probability calculations including conditional probability. Part (a) is routine construction, parts (b)(i)-(iii) involve basic probability rules and Bayes' theorem with clear given information, and part (c) adds a simple binomial element. All techniques are standard S3 material with no novel insight required, making it slightly easier than average. |
| 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 | Mark | Guidance |
| Tree diagram with first branches: On time (0.9), Late (0.1) | B1 | Correct first split |
| Second branches from On time: Early (0.15), On time (0.75), Late (0.10) | B1 | Correct probabilities on second set |
| Second branches from Late: On time (0.35), Late (0.65) | B1 | Fully labelled; all probabilities correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{early or on time}) = 0.9 \times 0.15 + 0.9 \times 0.75 + 0.1 \times 0.35\) | M1 | Summing correct product pairs |
| \(= 0.135 + 0.675 + 0.035 = 0.845\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{left on time} \mid \text{arrived on time}) = \frac{0.9 \times 0.75}{P(\text{arrived on time})}\) | M1 | Correct conditional probability structure |
| \(P(\text{arrived on time}) = 0.9 \times 0.75 + 0.1 \times 0.35 = 0.675 + 0.035 = 0.71\) | A1 | |
| \(= \frac{0.675}{0.71} = 0.9507...\) | A1 | Accept \(\frac{675}{710}\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{not late arriving}) = 0.845\) (from (b)(i)) | ||
| \(P(\text{left late and not late arriving}) = 0.1 \times 0.35 = 0.035\) | M1 | |
| \(P(\text{left late} \mid \text{not late arriving}) = \frac{0.035}{0.845}\) | M1 | Correct conditional structure |
| \(= 0.04142...\) | A1 | Accept equivalent fraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{arrive late}) = 1 - 0.845 = 0.155\) | B1 | |
| \(P(\text{left on time and arrive late}) = 0.9 \times 0.10 = 0.09\) | ||
| \(P(\text{left late and arrive late}) = 0.1 \times 0.65 = 0.065\) | ||
| Given two trains arrive late, \(P(\text{exactly one left on time})\) = \(\binom{2}{1}(0.09/0.155)(0.065/0.155)\) ... using conditional approach | M1 | |
| \(= 2 \times \frac{0.09}{0.155} \times \frac{0.065}{0.155}\) ... | M1 A1 | |
| \(= 2 \times \frac{0.09 \times 0.065}{0.155^2} = \frac{0.0117}{0.024025} \approx 0.4871\) | A1 | Accept equivalent decimal |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tree diagram with first branches: On time (0.9), Late (0.1) | B1 | Correct first split |
| Second branches from On time: Early (0.15), On time (0.75), Late (0.10) | B1 | Correct probabilities on second set |
| Second branches from Late: On time (0.35), Late (0.65) | B1 | Fully labelled; all probabilities correct |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{early or on time}) = 0.9 \times 0.15 + 0.9 \times 0.75 + 0.1 \times 0.35$ | M1 | Summing correct product pairs |
| $= 0.135 + 0.675 + 0.035 = 0.845$ | A1 | Correct answer |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{left on time} \mid \text{arrived on time}) = \frac{0.9 \times 0.75}{P(\text{arrived on time})}$ | M1 | Correct conditional probability structure |
| $P(\text{arrived on time}) = 0.9 \times 0.75 + 0.1 \times 0.35 = 0.675 + 0.035 = 0.71$ | A1 | |
| $= \frac{0.675}{0.71} = 0.9507...$ | A1 | Accept $\frac{675}{710}$ or equivalent |
## Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{not late arriving}) = 0.845$ (from (b)(i)) | | |
| $P(\text{left late and not late arriving}) = 0.1 \times 0.35 = 0.035$ | M1 | |
| $P(\text{left late} \mid \text{not late arriving}) = \frac{0.035}{0.845}$ | M1 | Correct conditional structure |
| $= 0.04142...$ | A1 | Accept equivalent fraction |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{arrive late}) = 1 - 0.845 = 0.155$ | B1 | |
| $P(\text{left on time and arrive late}) = 0.9 \times 0.10 = 0.09$ | | |
| $P(\text{left late and arrive late}) = 0.1 \times 0.65 = 0.065$ | | |
| Given two trains arrive late, $P(\text{exactly one left on time})$ = $\binom{2}{1}(0.09/0.155)(0.065/0.155)$ ... using conditional approach | M1 | |
| $= 2 \times \frac{0.09}{0.155} \times \frac{0.065}{0.155}$ ... | M1 A1 | |
| $= 2 \times \frac{0.09 \times 0.065}{0.155^2} = \frac{0.0117}{0.024025} \approx 0.4871$ | A1 | Accept equivalent decimal |2 On a rail route between two stations, A and $\mathrm { B } , 90 \%$ of trains leave A on time and $10 \%$ of trains leave A late.
Of those trains that leave A on time, $15 \%$ arrive at B early, $75 \%$ arrive on time and $10 \%$ arrive late.
Of those trains that leave A late, $35 \%$ arrive at B on time and $65 \%$ arrive late.
\begin{enumerate}[label=(\alph*)]
\item Represent this information by a fully-labelled tree diagram.
\item Hence, or otherwise, calculate the probability that a train:
\begin{enumerate}[label=(\roman*)]
\item arrives at B early or on time;
\item left A on time, given that it arrived at B on time;
\item left A late, given that it was not late in arriving at B .
\end{enumerate}\item Two trains arrive late at B. Assuming that their journey times are independent, calculate the probability that exactly one train left A on time.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2013 Q2 [14]}}