| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two independent categorical choices |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question involving basic probability tables, conditional probability, and independence. Part (a) requires simple arithmetic to complete a probability table and identify mutually exclusive events. Part (b) applies standard conditional probability formulas with given values. Part (c) involves interpreting and calculating a compound probability using independence. All techniques are routine for S1 with no novel problem-solving required, making it easier than average but not trivial due to the multi-part structure and careful bookkeeping needed. |
| 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 |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | \(\boldsymbol { M }\) | \(\boldsymbol { M } ^ { \prime }\) | Total |
| \(\boldsymbol { E }\) | 0.16 | 0.28 | |
| \(\boldsymbol { E } ^ { \prime }\) | |||
| Total | 0.60 | 1.00 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \cap E) = 0.16\), \(P(M' \cap E) = 0.28 - 0.16 = 0.12\) | B1 | |
| \(P(M) = 1 - 0.60 = 0.40\) | B1 | |
| Remaining cells: \(P(M \cap E') = 0.24\), \(P(M' \cap E') = 0.36\), \(P(E') = 0.72\), \(P(E) = 0.28\) | B1 | All values correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(M\) | \(M'\) | Total |
| \(E\) | 0.16 | 0.12 |
| \(E'\) | 0.24 | 0.36 |
| Total | 0.40 | 0.60 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \cap E') + P(M' \cap E)\) | M1 | Correct method |
| \(= 0.24 + 0.12 = 0.36\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \cap E) = 0.16 \neq 0\) | B1 | Must give numerical value |
| Therefore they can both occur simultaneously, so not mutually exclusive | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \cap E) \times P(S \cap T)\) | M1 | Need \(P(S \cap T) = P(S) \times P(T\ |
| \(P(T\ | S) = 0.20\), so \(P(S \cap T) = 0.85 \times 0.20 = 0.17\) | |
| \(= 0.16 \times 0.17 = 0.0272\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M' \cap E') \times P(S' \cap T')\) | M1 | |
| \(P(S') = 0.15\), \(P(T'\ | S') = 0.25\), so \(P(S' \cap T') = 0.15 \times 0.25 = 0.0375\) | |
| \(= 0.36 \times 0.0375 = 0.0135\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Chris buys a morning newspaper (only) on Friday, and buys a morning newspaper on Saturday but not on Sunday | B2 | B1 for each correct component; must reference context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \cap E') \times P(S \cap T')\) | M1 | |
| \(P(S \cap T') = P(S) \times P(T'\ | S) = 0.85 \times 0.80 = 0.68\) | |
| \(= 0.24 \times 0.68 = 0.1632\) | A1 |
# Question 4:
## Part (a)(i) – Complete the probability table [3 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \cap E) = 0.16$, $P(M' \cap E) = 0.28 - 0.16 = 0.12$ | B1 | |
| $P(M) = 1 - 0.60 = 0.40$ | B1 | |
| Remaining cells: $P(M \cap E') = 0.24$, $P(M' \cap E') = 0.36$, $P(E') = 0.72$, $P(E) = 0.28$ | B1 | All values correct |
**Completed table:**
| | $M$ | $M'$ | Total |
|---|---|---|---|
| $E$ | 0.16 | 0.12 | 0.28 |
| $E'$ | 0.24 | 0.36 | 0.72 |
| Total | 0.40 | 0.60 | 1.00 |
## Part (a)(ii) – P(exactly one newspaper) [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \cap E') + P(M' \cap E)$ | M1 | Correct method |
| $= 0.24 + 0.12 = 0.36$ | A1 | |
## Part (a)(iii) – Justification M and E not mutually exclusive [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \cap E) = 0.16 \neq 0$ | B1 | Must give numerical value |
| Therefore they can both occur simultaneously, so not mutually exclusive | B1 | |
## Part (b)(i) – P(all four newspapers) [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \cap E) \times P(S \cap T)$ | M1 | Need $P(S \cap T) = P(S) \times P(T\|S)$ |
| $P(T\|S) = 0.20$, so $P(S \cap T) = 0.85 \times 0.20 = 0.17$ | | |
| $= 0.16 \times 0.17 = 0.0272$ | A1 | |
## Part (b)(ii) – P(none of the four newspapers) [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M' \cap E') \times P(S' \cap T')$ | M1 | |
| $P(S') = 0.15$, $P(T'\|S') = 0.25$, so $P(S' \cap T') = 0.15 \times 0.25 = 0.0375$ | | |
| $= 0.36 \times 0.0375 = 0.0135$ | A1 | |
## Part (c)(i) – State event $M \cap E' \cap S \cap T'$ [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| Chris buys a morning newspaper (only) on Friday, and buys a morning newspaper on Saturday but not on Sunday | B2 | B1 for each correct component; must reference context |
## Part (c)(ii) – Calculate $P(M \cap E' \cap S \cap T')$ [2 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \cap E') \times P(S \cap T')$ | M1 | |
| $P(S \cap T') = P(S) \times P(T'\|S) = 0.85 \times 0.80 = 0.68$ | | |
| $= 0.24 \times 0.68 = 0.1632$ | A1 | |4
\begin{enumerate}[label=(\alph*)]
\item Chris shops at his local store on his way to and from work every Friday.\\
The event that he buys a morning newspaper is denoted by $M$, and the event that he buys an evening newspaper is denoted by $E$.
On any one Friday, Chris may buy neither, exactly one or both of these newspapers.
\begin{enumerate}[label=(\roman*)]
\item Complete the table of probabilities, printed on the opposite page, where $M ^ { \prime }$ and $E ^ { \prime }$ denote the events 'not $M$ ' and 'not $E$ ' respectively.
\item Hence, or otherwise, find the probability that, on any given Friday, Chris buys exactly one newspaper.
\item Give a numerical justification for the following statement.\\
'The events $M$ and $E$ are not mutually exclusive.'
\end{enumerate}\item The event that Chris buys a morning newspaper on Saturday is denoted by $S$, and the event that he buys a morning newspaper on the following day, Sunday, is denoted by $T$. The event that he buys a morning newspaper on both Saturday and Sunday is denoted by $S \cap T$.
Each combination of the events $S$ and $T$ is independent of any combination of the events $M$ and $E$. However, the events $S$ and $T$ are not independent, with
$$\mathrm { P } ( S ) = 0.85 , \quad \mathrm { P } ( T \mid S ) = 0.20 \quad \text { and } \quad \mathrm { P } \left( T \mid S ^ { \prime } \right) = 0.75$$
Find the probability that, on a particular Friday, Saturday and Sunday, Chris buys:
\begin{enumerate}[label=(\roman*)]
\item all four newspapers;
\item none of the four newspapers.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item State, as briefly as possible, in the context of the question, the event that is denoted by $M \cap E ^ { \prime } \cap S \cap T ^ { \prime }$.
\item Calculate the value of $\mathrm { P } \left( M \cap E ^ { \prime } \cap S \cap T ^ { \prime } \right)$.
\section*{Answer space for question 4}
(a)(i)
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & $\boldsymbol { M }$ & $\boldsymbol { M } ^ { \prime }$ & Total \\
\hline
$\boldsymbol { E }$ & 0.16 & & 0.28 \\
\hline
$\boldsymbol { E } ^ { \prime }$ & & & \\
\hline
Total & & 0.60 & 1.00 \\
\hline
\end{tabular}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{4c679380-894f-4d36-aec8-296b662058e2-11_2050_1707_687_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q4 [15]}}