| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.3 This is a straightforward two-way table probability question requiring only basic probability operations (reading from table, adding probabilities, and applying conditional probability formula P(A|B) = P(A∩B)/P(B)). All calculations are direct with no conceptual challenges—significantly easier than average A-level questions which typically require more problem-solving or integration of multiple techniques. |
| 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 |
| \multirow{2}{*}{} | Weather | |||
| Fine ( \(F\) ) | Dull ( \(D\) ) | Wet ( \(\boldsymbol { W }\) ) | ||
| \multirow{4}{*}{Activity} | Cycling ( \(\boldsymbol { C }\) ) | 0.30 | 0.10 | 0 |
| Gardening ( \(\boldsymbol { G }\) ) | 0.25 | 0.05 | 0 | |
| Shopping ( \(\boldsymbol { S }\) ) | 0 | 0.10 | 0.05 | |
| Relaxing ( \(\boldsymbol { R }\) ) | 0 | 0.05 | 0.10 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F \cap C) = 0.3\) or \(\frac{3}{10}\) or \(30\%\) | B1 | CAO \((0.3)\) |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(G \cup S) = 0.45\) or \(\frac{45}{100}\) or \(45\%\) | B1 | CAO \((0.45)\) |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(C \mid F) = \dfrac{0.3 \text{ or (i)}}{0.55}\) | M1 | |
| \(\dfrac{30}{55}\) or \(\dfrac{6}{11}\) or \((0.54\) to \(0.55)\) or \((54\%\) to \(55\%)\) | A1 | CAO \(\left(\dfrac{6}{11}\right)\); AWFW \((0.54545)\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(R' \mid D) = \dfrac{0.25 \text{ or } (0.30 - 0.05)}{0.30}\) | M1 | Correct numerator |
| M1 | Correct denominator | |
| \(\dfrac{25}{30}\) or \(\dfrac{5}{6}\) or \((0.83\) to \(0.834)\) or \((83\%\) to \(83.4\%)\) | A1 | CAO \(\left(\dfrac{5}{6}\right)\); AWFW \((0.83333)\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F \mid C') = \dfrac{0.25 \text{ or } (0.60 - 0.35)}{0.60}\) | M1 | Correct expression |
| \(\dfrac{25}{60}\) or \(\dfrac{5}{12}\) or \((0.416\) to \(0.42)\) or \((41.6\%\) to \(42\%)\) | A1 | CAO \(\left(\dfrac{5}{12}\right)\); AWRT \((0.41667)\) |
| Total | (2, 3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = [P(F \cap C)]^2 + [P(F \cap G)]^2\) | M1 | Attempt at sum of at least 2 squared terms; \(0 < \text{term} < 1\); not \((a+b)^2\); may be implied by correct expression or correct answer |
| \(0.30^2 + 0.25^2\) or \(0.09 + 0.0625\) | A1 | OE; ignore additional terms or integer multipliers; may be implied by correct answer |
| \(\dfrac{1525}{10000}\) or \(\dfrac{305}{2000}\) or \(\dfrac{61}{400}\) or \((0.152\) to \(0.153)\) or \((15.2\%\) to \(15.3\%)\) | A1 | CAO \((0.1525)\); AWFW |
| Total | 3 |
# Question 5:
## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F \cap C) = 0.3$ or $\frac{3}{10}$ or $30\%$ | B1 | CAO $(0.3)$ |
| **Total** | **1** | |
## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(G \cup S) = 0.45$ or $\frac{45}{100}$ or $45\%$ | B1 | CAO $(0.45)$ |
| **Total** | **1** | |
## Part (a)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(C \mid F) = \dfrac{0.3 \text{ or (i)}}{0.55}$ | M1 | |
| $\dfrac{30}{55}$ or $\dfrac{6}{11}$ **or** $(0.54$ to $0.55)$ or $(54\%$ to $55\%)$ | A1 | CAO $\left(\dfrac{6}{11}\right)$; AWFW $(0.54545)$ |
| **Total** | **2** | |
## Part (a)(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(R' \mid D) = \dfrac{0.25 \text{ or } (0.30 - 0.05)}{0.30}$ | M1 | Correct numerator |
| | M1 | Correct denominator |
| $\dfrac{25}{30}$ or $\dfrac{5}{6}$ **or** $(0.83$ to $0.834)$ or $(83\%$ to $83.4\%)$ | A1 | CAO $\left(\dfrac{5}{6}\right)$; AWFW $(0.83333)$ |
| **Total** | **3** | |
## Part (a)(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F \mid C') = \dfrac{0.25 \text{ or } (0.60 - 0.35)}{0.60}$ | M1 | Correct expression |
| $\dfrac{25}{60}$ or $\dfrac{5}{12}$ **or** $(0.416$ to $0.42)$ or $(41.6\%$ to $42\%)$ | A1 | CAO $\left(\dfrac{5}{12}\right)$; AWRT $(0.41667)$ |
| **Total** | **(2, 3)** | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = [P(F \cap C)]^2 + [P(F \cap G)]^2$ | M1 | Attempt at sum of at least 2 squared terms; $0 < \text{term} < 1$; not $(a+b)^2$; may be implied by correct expression or correct answer |
| $0.30^2 + 0.25^2$ **or** $0.09 + 0.0625$ | A1 | OE; ignore additional terms or integer multipliers; may be implied by correct answer |
| $\dfrac{1525}{10000}$ or $\dfrac{305}{2000}$ or $\dfrac{61}{400}$ **or** $(0.152$ to $0.153)$ or $(15.2\%$ to $15.3\%)$ | A1 | CAO $(0.1525)$; AWFW |
| **Total** | **3** | |
---5 Roger is an active retired lecturer. Each day after breakfast, he decides whether the weather for that day is going to be fine ( $F$ ), dull ( $D$ ) or wet ( $W$ ). He then decides on only one of four activities for the day: cycling ( $C$ ), gardening ( $G$ ), shopping ( $S$ ) or relaxing $( R )$. His decisions from day to day may be assumed to be independent.
The table shows Roger's probabilities for each combination of weather and activity.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Weather} \\
\hline
& & Fine ( $F$ ) & Dull ( $D$ ) & Wet ( $\boldsymbol { W }$ ) \\
\hline
\multirow{4}{*}{Activity} & Cycling ( $\boldsymbol { C }$ ) & 0.30 & 0.10 & 0 \\
\hline
& Gardening ( $\boldsymbol { G }$ ) & 0.25 & 0.05 & 0 \\
\hline
& Shopping ( $\boldsymbol { S }$ ) & 0 & 0.10 & 0.05 \\
\hline
& Relaxing ( $\boldsymbol { R }$ ) & 0 & 0.05 & 0.10 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that, on a particular day, Roger decided:
\begin{enumerate}[label=(\roman*)]
\item that it was going to be fine and that he would go cycling;
\item on either gardening or shopping;
\item to go cycling, given that he had decided that it was going to be fine;
\item not to relax, given that he had decided that it was going to be dull;
\item that it was going to be fine, given that he did not go cycling.
\end{enumerate}\item Calculate the probability that, on a particular Saturday and Sunday, Roger decided that it was going to be fine and decided on the same activity for both days.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q5 [12]}}