| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Basic probability calculation |
| Difficulty | Easy -1.3 This is a straightforward application of basic probability rules (addition rule, independence) with clearly stated probabilities and standard multi-part structure. All parts require direct formula application with no problem-solving insight—typical S1 bookwork that's easier than average A-level maths questions. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A \cup M) = 0.70 + 0.55 - 0.45 = 0.80\) | M1 A1 | Use of addition formula |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{either but not both}) = 0.80 - 0.45 = 0.35\) | A1 | Follow through from (a)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| If independent: \(P(A) \times P(M) = 0.70 \times 0.55 = 0.385\) | M1 | Correct multiplication |
| \(0.385 \neq 0.45 = P(A \cap M)\), therefore not independent | A1 | Correct comparison and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A \cap M \cap B \cap N) = 0.70 \times 0.55 \times 0.85 \times 0.65\) | M1 | All four probabilities multiplied |
| \(= 0.213\) (3 s.f.) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{none attend}) = (1-0.70)(1-0.55)(1-0.85)(1-0.65)\) | M1 | All four complements identified |
| \(= 0.30 \times 0.45 \times 0.15 \times 0.35\) | M1 | Correct multiplication |
| \(= 0.00709\) (3 s.f.) | A1 | Correct answer |
## Question 4:
### Part (a)(i):
$P(A \cup M) = 0.70 + 0.55 - 0.45 = 0.80$ | M1 A1 | Use of addition formula
### Part (a)(ii):
$P(\text{either but not both}) = 0.80 - 0.45 = 0.35$ | A1 | Follow through from (a)(i)
### Part (b):
If independent: $P(A) \times P(M) = 0.70 \times 0.55 = 0.385$ | M1 | Correct multiplication
$0.385 \neq 0.45 = P(A \cap M)$, therefore not independent | A1 | Correct comparison and conclusion
### Part (c)(i):
$P(A \cap M \cap B \cap N) = 0.70 \times 0.55 \times 0.85 \times 0.65$ | M1 | All four probabilities multiplied
$= 0.213$ (3 s.f.) | A1 | Correct answer
### Part (c)(ii):
$P(\text{none attend}) = (1-0.70)(1-0.55)(1-0.85)(1-0.65)$ | M1 | All four complements identified
$= 0.30 \times 0.45 \times 0.15 \times 0.35$ | M1 | Correct multiplication
$= 0.00709$ (3 s.f.) | A1 | Correct answer
---4 Alf and Mabel are members of a bowls club and sometimes attend the club's social events.
The probability, $\mathrm { P } ( A )$, that Alf attends a social event is 0.70 .\\
The probability, $\mathrm { P } ( M )$, that Mabel attends a social event is 0.55 .\\
The probability, $\mathrm { P } ( A \cap M )$, that both Alf and Mabel attend the same social event is 0.45 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that:
\begin{enumerate}[label=(\roman*)]
\item either Alf or Mabel or both attend a particular social event;
\item either Alf or Mabel but not both attend a particular social event.
\end{enumerate}\item Give a numerical justification for the following statement.\\
"Events $A$ and $M$ are not independent."
\item Ben and Nora are also members of the bowls club and sometimes attend the club's social events.
The probability, $\mathrm { P } ( B )$, that Ben attends a social event is 0.85 .\\
The probability, $\mathrm { P } ( N )$, that Nora attends a social event is 0.65 .\\
The attendance of each of Ben and Nora at a social event is independent of the attendance of all other members.
Find the probability that:
\begin{enumerate}[label=(\roman*)]
\item all four named members attend a particular social event;
\item none of the four named members attend a particular social event.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q4 [10]}}