| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Sequential events and tree diagrams |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability table completion and conditional probability calculations. Part (a) requires simple arithmetic to complete the table, parts (b) and (c) involve reading values and applying the conditional probability formula P(A|B) = P(A∩B)/P(B), and part (d) uses the multinomial coefficient with given probabilities. All techniques are standard textbook exercises with no novel insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| \multirow{2}{*}{} | Arrive at A | ||||
| Early | On time | Late | Total | ||
| \multirow{2}{*}{Depart from D} | On time | 0.16 | 0.56 | 0.08 | |
| Late | |||||
| Total | 0.22 | 0.65 | 1.00 | ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| On time total \(= 0.16 + 0.56 + 0.08 = 0.80\) | B1 | |
| Late row: Early \(= 0.06\), On time \(= 0.09\), Late \(= ?\) | ||
| Late total \(= 0.20\); Late arriving late \(= 0.13\) | B1 | Both values in Late row correct; table complete |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{departs on time and arrives on time}) = 0.56\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{departs late}) = 0.20\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{late} \mid \text{departed late}) = \dfrac{P(\text{late and departed late})}{P(\text{departed late})}\) | M1 | Correct conditional probability method |
| \(= \dfrac{0.13}{0.20} = 0.65\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{not late} \mid \text{departed on time}) = \dfrac{P(\text{not late and departed on time})}{P(\text{departed on time})}\) | M1 | |
| \(= \dfrac{0.16 + 0.56}{0.80} = \dfrac{0.72}{0.80} = 0.90\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probabilities when departed on time: \(P(E)=0.20\), \(P(O)=0.70\), \(P(L)=0.10\) | B1 | Correct conditional probabilities identified |
| \(P(\text{early once, on time once, late once}) = 3! \times 0.20 \times 0.70 \times 0.10\) | M1 | Multinomial with \(3!\) |
| \(= 6 \times 0.014 = 0.084\) | A1A1 | Awrt \(0.084\) |
# Question 3:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| On time total $= 0.16 + 0.56 + 0.08 = 0.80$ | B1 | |
| Late row: Early $= 0.06$, On time $= 0.09$, Late $= ?$ | | |
| Late total $= 0.20$; Late arriving late $= 0.13$ | B1 | Both values in Late row correct; table complete |
## Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{departs on time and arrives on time}) = 0.56$ | B1 | |
## Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{departs late}) = 0.20$ | B1 | |
## Part (c)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{late} \mid \text{departed late}) = \dfrac{P(\text{late and departed late})}{P(\text{departed late})}$ | M1 | Correct conditional probability method |
| $= \dfrac{0.13}{0.20} = 0.65$ | A1 | |
## Part (c)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{not late} \mid \text{departed on time}) = \dfrac{P(\text{not late and departed on time})}{P(\text{departed on time})}$ | M1 | |
| $= \dfrac{0.16 + 0.56}{0.80} = \dfrac{0.72}{0.80} = 0.90$ | A1 | |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probabilities when departed on time: $P(E)=0.20$, $P(O)=0.70$, $P(L)=0.10$ | B1 | Correct conditional probabilities identified |
| $P(\text{early once, on time once, late once}) = 3! \times 0.20 \times 0.70 \times 0.10$ | M1 | Multinomial with $3!$ |
| $= 6 \times 0.014 = 0.084$ | A1A1 | Awrt $0.084$ |
I can see these are answer space pages (pages 7-11) from what appears to be a statistics exam paper (P/Jun15/MS1B). The pages shown contain only blank answer spaces and the question paper for Question 4.
No mark scheme content is visible in these images — they show only the student answer booklet pages and the question text itself. To extract mark scheme content, I would need to see the actual mark scheme document.
From the question paper content visible, here is a summary of **Question 4** as asked (not mark scheme):3 A ferry sails once each day from port D to port A. The ferry departs from D on time or late but never early. However, the ferry can arrive at A early, on time or late.
The probabilities for some combined events of departing from $D$ and arriving at $A$ are shown in the table below.
\begin{enumerate}[label=(\alph*)]
\item Complete the table.
\item Write down the probability that, on a particular day, the ferry:
\begin{enumerate}[label=(\roman*)]
\item both departs and arrives on time;
\item departs late.
\end{enumerate}\item Find the probability that, on a particular day, the ferry:
\begin{enumerate}[label=(\roman*)]
\item arrives late, given that it departed late;
\item does not arrive late, given that it departed on time.
\end{enumerate}\item On three particular days, the ferry departs from port D on time.
Find the probability that, on these three days, the ferry arrives at port A early once, on time once and late once. Give your answer to three decimal places.\\[0pt]
[4 marks]\\
(a)
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Answer space for question 3}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Arrive at A} & \\
\hline
& & Early & On time & Late & Total \\
\hline
\multirow{2}{*}{Depart from D} & On time & 0.16 & 0.56 & 0.08 & \\
\hline
& Late & & & & \\
\hline
& Total & 0.22 & 0.65 & & 1.00 \\
\hline
\end{tabular}
\end{center}
\end{table}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q3 [13]}}