| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate combined outcome probability |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question with clearly stated probabilities requiring basic probability rules (complement, addition) and the multiplication rule for independent events. Part (a) involves simple arithmetic with given probabilities, while part (b) applies the law of total probability and basic binomial probability—all standard S1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(C') = 1 - P(C) = 1 - 0.6 = 0.4\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(C \cap B') = 0.6 - 0.25 = 1 - (0.4 + 0.25) = 0.35\) | M1 (A1) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B) = (i) + p\) with \(p < 0.6 = (i) + 0.25 = 0.65\) OR \(P(B) = 1 - (ii) = 0.65\) OR \(1 = P(C) + P(B) - P(C \cap B)\); Thus \(P(B) = 1 - (0.6 - 0.25) = 0.65\) | M1 (A1) (A1) or (M2) (A1) or (M1) (A1) (A1) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(G \cap L) ⟹ (a)(ii) \times 0.9 = 0.25 \times 0.7 = (0.315) + (0.175) + [(a)(iii) - 0.25] \times 0.3 = (0.12)\) Note: Each pair of multiplied probabilities must be > 0 to score the corresponding method mark \(⟹ 0.315 + 0.175 + 0.12 = 0.61\) | M1 (M1) (M1) (A1) | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Probability = \([1 - (b)(i)]^5 = 0.39^5 = 0.009\) | M1 (A1) | 2 |
**4(a)(i)**
| $P(C') = 1 - P(C) = 1 - 0.6 = 0.4$ | B1 | 1 | CAO; or equivalent |
**4(a)(ii)**
| $P(C \cap B') = 0.6 - 0.25 = 1 - (0.4 + 0.25) = 0.35$ | M1 (A1) | 2 | Can be implied by correct answer; CAO; or equivalent |
**4(a)(iii)**
| $P(B) = (i) + p$ with $p < 0.6 = (i) + 0.25 = 0.65$ OR $P(B) = 1 - (ii) = 0.65$ OR $1 = P(C) + P(B) - P(C \cap B)$; Thus $P(B) = 1 - (0.6 - 0.25) = 0.65$ | M1 (A1) (A1) or (M2) (A1) or (M1) (A1) (A1) | 3 | Can be implied by correct answer; Can be implied by correct answer; CAO; or equivalent |
**4(b)(i)**
| $P(G \cap L) ⟹ (a)(ii) \times 0.9 = 0.25 \times 0.7 = (0.315) + (0.175) + [(a)(iii) - 0.25] \times 0.3 = (0.12)$ Note: Each pair of multiplied probabilities must be > 0 to score the corresponding method mark $⟹ 0.315 + 0.175 + 0.12 = 0.61$ | M1 (M1) (M1) (A1) | 4 | Follow through or correct; Can be implied by correct answer; Follow through or correct; Ignore any multiplying factors; Ignore any additional terms; CAO |
**4(b)(ii)**
| Probability = $[1 - (b)(i)]^5 = 0.39^5 = 0.009$ | M1 (A1) | 2 | Allow $5 \times [1 - (b)(i)]^5$; AWRT (0.00902) |
**Total for Q4: 12 marks**
---4 Gary and his neighbour Larry work at the same place.\\
On any day when Gary travels to work, he uses one of three options: his car only, a bus only or both his car and a bus. The probability that he uses his car, either on its own or with a bus, is 0.6 . The probability that he uses both his car and a bus is 0.25 .
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that, on any particular day when Gary travels to work, he:
\begin{enumerate}[label=(\roman*)]
\item does not use his car;
\item uses his car only;
\item uses a bus.
\end{enumerate}\item On any day, the probability that Larry travels to work with Gary is 0.9 when Gary uses his car only, is 0.7 when Gary uses both his car and a bus, and is 0.3 when Gary uses a bus only.
\begin{enumerate}[label=(\roman*)]
\item Calculate the probability that, on any particular day when Gary travels to work, Larry travels with him.
\item Assuming that option choices are independent from day to day, calculate, to three decimal places, the probability that, during any particular week (5 days) when Gary travels to work every day, Larry never travels with him.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2009 Q4 [12]}}