| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Tree diagram with two-stage events |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question requiring only basic probability rules (independence and conditional probability). Part (a) involves simple multiplication of independent probabilities, while part (b) applies the definition of conditional probability with clear given values. All calculations are routine with no problem-solving insight required, making it easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | ||
| \(P(X) = 0.3\), \(P(Y) = 0.4\), \(P(Z) = 0.2\) | ||
| 2(a)(i) | ||
| \(P(X \cap Y \cap Z) = 0.3 \times 0.4 \times 0.2 = 0.024\) | M1 | 1 |
| 2(a)(ii) | ||
| \(P(X' \cap Y' \cap Z') = 0.7 \times 0.6 \times 0.8 = 0.336\) | M1, A1 | At least 2 correct terms, CAO |
| 2 | ||
| 2(a)(iii) | ||
| \(P(X' \cap Y' \cap Z) = 0.7 \times 0.6 \times 0.2 = 0.084\) | M1, A1 | Correct numerical expression, CAO |
| 2(b) | ||
| \(P(W \mid Z) = 0.9\), \(P(W \mid Z') = 0.25\) | ||
| 2(b)(i) | ||
| \(P(Z \cap W) = 0.2 \times 0.9 = 0.18\) | M1, A1 | Correct numerical expression, CAO |
| 2 | ||
| 2(b)(ii) | ||
| \(P((Z \cap W) \cup (Z' \cap W))\) or \(1 - [P(Z \cap W) \cup (Z' \cap W))]\) | ||
| \(= 0.2 \times (1 - 0.9) + (1 - 0.2) \times 0.25\) | M1 | \(0.2 \times 0.9\) or (b)(i) |
| M1 | \((1 - 0.2) \times (1 - 0.25)\) | |
| Cannot score an M1 in both methods | ||
| \(= 0.02 + 0.20 = 0.22\) | A1 | \(1 - (0.18 + 0.60)\), CAO |
| 3 |
| **2(a)** |
|----------|
| $P(X) = 0.3$, $P(Y) = 0.4$, $P(Z) = 0.2$ | | |
| **2(a)(i)** |
|----------|
| $P(X \cap Y \cap Z) = 0.3 \times 0.4 \times 0.2 = 0.024$ | M1 | 1 |
| **2(a)(ii)** |
|----------|
| $P(X' \cap Y' \cap Z') = 0.7 \times 0.6 \times 0.8 = 0.336$ | M1, A1 | At least 2 correct terms, CAO |
| | | 2 |
| **2(a)(iii)** |
|----------|
| $P(X' \cap Y' \cap Z) = 0.7 \times 0.6 \times 0.2 = 0.084$ | M1, A1 | Correct numerical expression, CAO |
| | | |
| **2(b)** |
|----------|
| $P(W \mid Z) = 0.9$, $P(W \mid Z') = 0.25$ | | |
| **2(b)(i)** |
|----------|
| $P(Z \cap W) = 0.2 \times 0.9 = 0.18$ | M1, A1 | Correct numerical expression, CAO |
| | | 2 |
| **2(b)(ii)** |
|----------|
| $P((Z \cap W) \cup (Z' \cap W))$ or $1 - [P(Z \cap W) \cup (Z' \cap W))]$ | | |
| $= 0.2 \times (1 - 0.9) + (1 - 0.2) \times 0.25$ | M1 | $0.2 \times 0.9$ or (b)(i) |
| | M1 | $(1 - 0.2) \times (1 - 0.25)$ |
| | | Cannot score an M1 in both methods |
| $= 0.02 + 0.20 = 0.22$ | A1 | $1 - (0.18 + 0.60)$, CAO |
| | | 3 |
**Total for Q2: 11**
---2 Xavier, Yuri and Zara attend a sports centre for their judo club's practice sessions. The probabilities of them arriving late are, independently, $0.3,0.4$ and 0.2 respectively.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that for a particular practice session:
\begin{enumerate}[label=(\roman*)]
\item all three arrive late;
\item none of the three arrives late;
\item only Zara arrives late.
\end{enumerate}\item Zara's friend, Wei, also attends the club's practice sessions. The probability that Wei arrives late is 0.9 when Zara arrives late, and is 0.25 when Zara does not arrive late.
Calculate the probability that for a particular practice session:
\begin{enumerate}[label=(\roman*)]
\item both Zara and Wei arrive late;
\item either Zara or Wei, but not both, arrives late.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2006 Q2 [10]}}