| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Three or more stages |
| Difficulty | Moderate -0.3 This is a standard S3 conditional probability question using tree diagrams and the law of total probability. Part (a) is routine diagram construction, part (b) involves straightforward applications of probability rules with clearly defined conditional probabilities, and part (c) is a basic binomial probability calculation. While it requires careful bookkeeping across multiple stages, it demands no novel insight—just systematic application of A-level probability formulas to a multi-stage scenario. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing tree with structure: 2 × 2 × 3 = 12 branches | M1 | Shape; 2 × 2 × 3 = 12 branches |
| Correct labels OT & L and E & OT & L | M1 | Labels; OT & L and E & OT & L |
| Attempt at percentages or probabilities for D and M and T | M1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T_{OT}) = 0.351 + 0.063 + 0.009 + 0.017 = \mathbf{0.44}\) | B1 | CAO |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T_{OT} \mid D_{OT}) = \frac{0.351 + 0.063}{0.9} = \frac{0.414}{0.9} = \mathbf{0.46}\) | M1 | Correct numerator; PI |
| A1 | CAO | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T_{E \text{ or } OT} \mid D_{OT}) = 0.46 + \frac{0.14625 + 0.00315}{0.9} =\) | M1 | (ii) + p |
| \(0.46 + \frac{0.17775}{0.9} = 0.46 + \mathbf{0.197 \text{ to } 0.20}\) | A1 | AWFW; PI (0.1975) |
| \(= \mathbf{0.657 \text{ to } 0.66}\) | A1 | AWFW (0.6575) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T_{E \text{ or } OT} \mid M_{OT}) = \frac{0.14625 + 0.351 + 0.00375 + 0.009}{0.9 \times 0.65 + 0.1 \times 0.15} = \frac{0.51}{0.6} = \mathbf{0.85}\) | M1 | Correct numerator; PI |
| A1 | CAO | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 \times 0.60 = 0.85 \Rightarrow B2\) | \(2 \times 1 - 0.15 = 0.85 \Rightarrow B2\) | |
| 8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T_{OT} \mid D_{OT}) = 0.46\) | B1 | AWFW; PI (0.1975) |
| \(P(T_E \mid D_{OT}) = 0.6575 - 0.46 = \mathbf{0.197 \text{ to } 0.20}\) | M1 | \(p_1^2 \times p_2\) |
| \(P(T_{OT} \cap T_{OT} \cap T_E) = 0.46^2 \times 0.1975\) | ml | CAO |
| \(\times 3\) | A1 | AWFW (0.12537) |
| \(= \mathbf{0.125 \text{ to } 0.126}\) | ||
| 4 | ||
| Total | 15 |
## Part (a)
Diagram showing tree with structure: 2 × 2 × 3 = 12 branches | M1 | Shape; 2 × 2 × 3 = 12 branches
Correct labels OT & L and E & OT & L | M1 | Labels; OT & L and E & OT & L
Attempt at percentages or probabilities for D and M and T | M1 |
| | **3** |
## Part (b)(i)
$P(T_{OT}) = 0.351 + 0.063 + 0.009 + 0.017 = \mathbf{0.44}$ | B1 | CAO
| | **1** |
## Part (b)(ii)
$P(T_{OT} \mid D_{OT}) = \frac{0.351 + 0.063}{0.9} = \frac{0.414}{0.9} = \mathbf{0.46}$ | M1 | Correct numerator; PI
| A1 | CAO
| | **2** |
## Part (b)(iii)
$P(T_{E \text{ or } OT} \mid D_{OT}) = 0.46 + \frac{0.14625 + 0.00315}{0.9} =$ | M1 | (ii) + p
$0.46 + \frac{0.17775}{0.9} = 0.46 + \mathbf{0.197 \text{ to } 0.20}$ | A1 | AWFW; PI (0.1975)
$= \mathbf{0.657 \text{ to } 0.66}$ | A1 | AWFW (0.6575)
| | **3** |
## Part (b)(iv)
$P(T_{E \text{ or } OT} \mid M_{OT}) = \frac{0.14625 + 0.351 + 0.00375 + 0.009}{0.9 \times 0.65 + 0.1 \times 0.15} = \frac{0.51}{0.6} = \mathbf{0.85}$ | M1 | Correct numerator; PI
| A1 | CAO
| | **2** |
## Special Cases
$1 \times 0.60 = 0.85 \Rightarrow B2$ | | $2 \times 1 - 0.15 = 0.85 \Rightarrow B2$
| | **8** |
## Part (c)
$P(T_{OT} \mid D_{OT}) = 0.46$ | B1 | AWFW; PI (0.1975)
$P(T_E \mid D_{OT}) = 0.6575 - 0.46 = \mathbf{0.197 \text{ to } 0.20}$ | M1 | $p_1^2 \times p_2$
$P(T_{OT} \cap T_{OT} \cap T_E) = 0.46^2 \times 0.1975$ | ml | CAO
$\times 3$ | A1 | AWFW (0.12537)
$= \mathbf{0.125 \text{ to } 0.126}$ | |
| | **4** |
**Total** | | **15** |
---A plane flies regularly between airports D and T with an intermediate stop at airport M. The time of the plane's departure from, or arrival at, each airport is classified as either early, on time, or late.
On 90\% of flights, the plane departs from D on time, and on 10\% of flights, it departs from D late.
Of those flights that depart from D on time, 65\% then depart from M on time and 35\% depart from M late.
Of those flights that depart from D late, 15\% then depart from M on time and 85\% depart from M late.
Any flight that departs from M on time has probability 0.25 of arriving at T early, probability 0.60 of arriving at T on time and probability 0.15 of arriving at T late.
Any flight that departs from M late has probability 0.10 of arriving at T early, probability 0.20 of arriving at T on time and probability 0.70 of arriving at T late.
\begin{enumerate}[label=(\alph*)]
\item Represent this information by a tree diagram on which labels and percentages or probabilities are shown. [3 marks]
\item Hence, or otherwise, calculate the probability that the plane:
\begin{enumerate}[label=(\roman*)]
\item arrives at T on time;
\item arrives at T on time, given that it departed from D on time;
\item does not arrive at T late, given that it departed from D on time;
\item does not arrive at T late, given that it departed from M on time.
\end{enumerate}
[8 marks]
\item Three independent flights of the plane depart from D on time.
Calculate the probability that two flights arrive at T on time and that one flight arrives at T early. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2016 Q2 [15]}}