| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Multi-stage game or match outcomes |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question using tree diagrams with clearly stated probabilities. Part (i) requires basic application of Bayes' theorem or tree diagram calculation, while part (ii) involves summing three paths through a tree diagram. The structure is standard for S1 level with no conceptual surprises, making it slightly easier than average A-level maths. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i): \(P(W_1 | L_2) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.6}\) | B1 |
| B1 | For correct numerator | |
| M1 | For summing two factor products in denom | |
| A1 | For correct denominator unsimplified | |
| \(= \frac{0.18}{0.42} = 0.429\) | A1 | 5 |
| Part (ii): \(P(W_1 W_2 L_3) = 0.6 \times 0.7 \times 0.3 = 0.126\) | M1 | For summing three probability options |
| \(P(W_1 L_2 W_3) = 0.6 \times 0.3 \times 0.4 = 0.072\) | B1 | For one correct probability option |
| \(P(L_1 W_2 W_3) = 0.4 \times 0.4 \times 0.7 = 0.112\) | B1 | For two correct probability options |
| \(\text{Probability} = 0.31\) | A1 | 4 |
**Part (i):** $P(W_1 | L_2) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.6}$ | B1 | For $0.6 \times 0.3$ seen anywhere in isolation
| B1 | For correct numerator
| M1 | For summing two factor products in denom
| A1 | For correct denominator unsimplified
$= \frac{0.18}{0.42} = 0.429$ | A1 | 5 | For correct answer
**Part (ii):** $P(W_1 W_2 L_3) = 0.6 \times 0.7 \times 0.3 = 0.126$ | M1 | For summing three probability options
$P(W_1 L_2 W_3) = 0.6 \times 0.3 \times 0.4 = 0.072$ | B1 | For one correct probability option
$P(L_1 W_2 W_3) = 0.4 \times 0.4 \times 0.7 = 0.112$ | B1 | For two correct probability options
$\text{Probability} = 0.31$ | A1 | 4 | For correct answer
---5 Rachel and Anna play each other at badminton. Each game results in either a win for Rachel or a win for Anna. The probability of Rachel winning the first game is 0.6 . If Rachel wins a particular game, the probability of her winning the next game is 0.7 , but if she loses, the probability of her winning the next game is 0.4 . By using a tree diagram, or otherwise,\\
(i) find the conditional probability that Rachel wins the first game, given that she loses the second,\\
(ii) find the probability that Rachel wins 2 games and loses 1 game out of the first three games they play.
\hfill \mbox{\textit{CAIE S1 2002 Q5 [9]}}