| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Two-way table probabilities |
| Difficulty | Moderate -0.8 This is a straightforward two-way table probability question requiring only basic probability definitions and arithmetic. Students must read values from the table, calculate totals, and apply standard formulas for P(A), P(A'∩B), P(A∪B), test independence using P(A∩B)=P(A)P(B), and identify mutually exclusive events. All steps are routine with no problem-solving insight required, making it easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| \multirow{2}{*}{} | Number of toilets | ||
| One | Two | Three | |
| Terraced | 20 | 10 | 4 |
| Semi-Detached | 18 | 50 | 16 |
| Detached | 12 | 10 | 8 |
| Apartment | 22 | 30 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A) = \frac{100}{200} = 0.5\) | B1 | States correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A' \cap B) = \frac{20}{200} = 0.1\) | B1 | States correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B) = \frac{30}{200} = 0.15\) | M1 | Uses correct formula for \(P(A \cup B)\) with 'their' \(P(A)\), \(P(B)\) and \(P(A \cap B) \neq 0\). Or counting from table: \(\frac{12+8+10+50+30+10}{200}\). Condone double-counting the 10 for \(A \cap B\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A \cup B) = 0.5 + 0.15 - 0.05 = 0.6\) | A1 | Obtains correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A) \times P(B) = 0.5 \times 0.15 = 0.075\), \(0.075 \neq 0.05\) | M1 | Calculates \(P(A) \times P(B)\) for 'their' \(P(A)\) and \(P(B)\) |
| As \(P(A \cap B) \neq P(A) \times P(B)\), events \(A\) and \(B\) are not independent | A1F | Compares to 'their' \(P(A \cap B)\) (providing \(P(A \cap B) \neq 0\)) and deduces events are not independent |
| Answer | Marks | Guidance |
|---|---|---|
| Event 1 is the event the property has 1 toilet; Event 2 is the event the property has 3 toilets | B1 | States any two events which are mutually exclusive, but must be different to events \(A\) and \(B\). Accept: 2 from terraced/semi-detached/apartment; or 1 toilet and 3 toilets; or apartment and 3 toilets |
# Question 17(a)(i):
$P(A) = \frac{100}{200} = 0.5$ | B1 | States correct answer
---
# Question 17(a)(ii):
$P(A' \cap B) = \frac{20}{200} = 0.1$ | B1 | States correct answer
---
# Question 17(a)(iii):
$P(B) = \frac{30}{200} = 0.15$ | M1 | Uses correct formula for $P(A \cup B)$ with 'their' $P(A)$, $P(B)$ and $P(A \cap B) \neq 0$. Or counting from table: $\frac{12+8+10+50+30+10}{200}$. Condone double-counting the 10 for $A \cap B$.
$P(A \cap B) = \frac{10}{200} = 0.05$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.5 + 0.15 - 0.05 = 0.6$ | A1 | Obtains correct answer
---
# Question 17(b):
$P(A) \times P(B) = 0.5 \times 0.15 = 0.075$, $0.075 \neq 0.05$ | M1 | Calculates $P(A) \times P(B)$ for 'their' $P(A)$ and $P(B)$
As $P(A \cap B) \neq P(A) \times P(B)$, events $A$ and $B$ are not independent | A1F | Compares to 'their' $P(A \cap B)$ (providing $P(A \cap B) \neq 0$) and deduces events are not independent
---
# Question 17(c):
Event 1 is the event the property has 1 toilet; Event 2 is the event the property has 3 toilets | B1 | States any two events which are mutually exclusive, but must be different to events $A$ and $B$. Accept: 2 from terraced/semi-detached/apartment; or 1 toilet and 3 toilets; or apartment and 3 toilets
---17 The number of toilets in each of a random sample of 200 properties from a town was recorded.
Four types of properties were included: terraced, semi-detached, detached and apartment.
The data is summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multirow{2}{*}{} & \multicolumn{3}{|c|}{Number of toilets} \\
\hline
& One & Two & Three \\
\hline
Terraced & 20 & 10 & 4 \\
\hline
Semi-Detached & 18 & 50 & 16 \\
\hline
Detached & 12 & 10 & 8 \\
\hline
Apartment & 22 & 30 & 0 \\
\hline
\end{tabular}
\end{center}
One of the properties is selected at random.\\
$A$ is the event 'the property has exactly two toilets'.\\
$B$ is the event 'the property is detached'.\\
17
\begin{enumerate}[label=(\alph*)]
\item (i) Find $\mathrm { P } ( A )$.
17 (a) (ii) Find $\mathrm { P } \left( A ^ { \prime } \cap B \right)$.
17 (a) (iii) Find $\mathrm { P } ( A \cup B )$.\\
17
\item Determine whether events $A$ and $B$ are independent.\\
Fully justify your answer.\\
17
\item Using the table, write down two events, other than event $\boldsymbol { A }$ and event $\boldsymbol { B }$, which are mutually exclusive.
Event 1 $\_\_\_\_$
\section*{Event 2}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2021 Q17 [7]}}