| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Sequential events and tree diagrams |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using tree diagrams with clearly stated probabilities. Part (a) requires basic multiplication of probabilities along branches, while part (b) adds independent events and requires identifying multiple pathways for 'exactly 2'. All calculations are routine applications of probability rules with no conceptual challenges or novel problem-solving required. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(F \cap D) = P(F) \times P(D \mid F) = 0.8 \times 0.9\) | M1 | |
| \(= 0.72\) | A1 | CAO (18/25) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(F' \cap D') = P(F') \times P(D' \mid F') = (1-0.8) \times (1-0.4)\) | M1 | |
| \(= 0.2 \times 0.6 = 0.12\) | A1 | CAO (3/25) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(M) = 0.7\) | ||
| \(P(F \cap D \cap M) = P(F) \times P(D \mid F) \times P(M)\) | M1 | (a)(i) \(\times\) P(M); ignore multipliers |
| \(= \text{(a)(i)} \times P(M) = 0.72 \times 0.7\) | A1\(\sqrt{}\) | Or equivalent; \(\sqrt{}\) on (a)(i) \(< 1\) |
| \(= 0.504\) | A1 | CAO (63/125) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(2 \text{ in } 3) = P(F \cap D \cap M') + P(F \cap D' \cap M) + P(F' \cap D \cap M)\) | M1 | At least 2 permutations of 3 events seen, or implied by multiplication of 3 correct probabilities at least twice; ignore multipliers e.g. \(\times 3\) |
| \(= 0.8 \times 0.9 \times 0.3 + 0.8 \times 0.1 \times 0.7 + 0.2 \times 0.4 \times 0.7\) | A2 (A1) | At least 2 correct expressions (exactly 1 correct expression) |
| \(= 0.216 + 0.056 + 0.056\) | ||
| \(= 0.328\) | A1 | CAO (41/125) |
# Question 3:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(F \cap D) = P(F) \times P(D \mid F) = 0.8 \times 0.9$ | M1 | |
| $= 0.72$ | A1 | CAO (18/25) |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(F' \cap D') = P(F') \times P(D' \mid F') = (1-0.8) \times (1-0.4)$ | M1 | |
| $= 0.2 \times 0.6 = 0.12$ | A1 | CAO (3/25) |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(M) = 0.7$ | | |
| $P(F \cap D \cap M) = P(F) \times P(D \mid F) \times P(M)$ | M1 | (a)(i) $\times$ P(M); ignore multipliers |
| $= \text{(a)(i)} \times P(M) = 0.72 \times 0.7$ | A1$\sqrt{}$ | Or equivalent; $\sqrt{}$ on (a)(i) $< 1$ |
| $= 0.504$ | A1 | CAO (63/125) |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2 \text{ in } 3) = P(F \cap D \cap M') + P(F \cap D' \cap M) + P(F' \cap D \cap M)$ | M1 | At least 2 permutations of 3 events seen, or implied by multiplication of 3 correct probabilities at least twice; ignore multipliers e.g. $\times 3$ |
| $= 0.8 \times 0.9 \times 0.3 + 0.8 \times 0.1 \times 0.7 + 0.2 \times 0.4 \times 0.7$ | A2 (A1) | At least 2 correct expressions (exactly 1 correct expression) |
| $= 0.216 + 0.056 + 0.056$ | | |
| $= 0.328$ | A1 | CAO (41/125) |
---3 Fred and his daughter, Delia, support their town's rugby team. The probability that Fred watches a game is 0.8 . The probability that Delia watches a game is 0.9 when her father watches the game, and is 0.4 when her father does not watch the game.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that:
\begin{enumerate}[label=(\roman*)]
\item both Fred and Delia watch a particular game;
\item neither Fred nor Delia watch a particular game.
\end{enumerate}\item Molly supports the same rugby team as Fred and Delia. The probability that Molly watches a game is 0.7 , and is independent of whether or not Fred or Delia watches the game.
Calculate the probability that:
\begin{enumerate}[label=(\roman*)]
\item all 3 supporters watch a particular game;
\item exactly 2 of the 3 supporters watch a particular game.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q3 [11]}}