| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward application of Bayes' theorem with clearly stated probabilities and a standard tree diagram setup. Part (i) requires only organizing given information visually, and part (ii) is a direct calculation using P(A|B) = P(A∩B)/P(B) with no conceptual challenges—easier than average A-level work. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Tree diagram with correct shape: branches showing T(0.65), C(0.28), HC(0.07) from start; M(0.8), NM(0.2) after T; M(0.5), NM(0.5) after C; M(1), NM(0) after HC | M1 | Correct shape with either one branch after HC or 2 branches with 0 prob seen correct. Labelled and clear annotation |
| All probabilities correct | A1 [2] | All probs correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(C \mid \text{milk}) = \dfrac{P(\text{coffee} \cap \text{milk})}{P(\text{milk})}\) | M1 | Attempt at \(P(\text{coffee} \cap \text{milk})\) as a two-factor product only seen as numerator or denominator of a fraction |
| \(= \dfrac{0.28 \times 0.5}{0.65 \times 0.8 + 0.28 \times 0.5 + 0.07 \times 1}\) | M1 | Summing appropriate three 2-factor products seen anywhere (can omit the 1) |
| \(= \dfrac{0.14}{0.73} = 0.192\) | A1 [3] | Correct answer oe |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram with correct shape: branches showing T(0.65), C(0.28), HC(0.07) from start; M(0.8), NM(0.2) after T; M(0.5), NM(0.5) after C; M(1), NM(0) after HC | M1 | Correct shape with either one branch after HC or 2 branches with 0 prob seen correct. Labelled and clear annotation |
| All probabilities correct | A1 [2] | All probs correct |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(C \mid \text{milk}) = \dfrac{P(\text{coffee} \cap \text{milk})}{P(\text{milk})}$ | M1 | Attempt at $P(\text{coffee} \cap \text{milk})$ as a two-factor product only seen as numerator or denominator of a fraction |
| $= \dfrac{0.28 \times 0.5}{0.65 \times 0.8 + 0.28 \times 0.5 + 0.07 \times 1}$ | M1 | Summing appropriate three 2-factor products seen anywhere (can omit the 1) |
| $= \dfrac{0.14}{0.73} = 0.192$ | A1 [3] | Correct answer oe |
---1 Ayman's breakfast drink is tea, coffee or hot chocolate with probabilities $0.65,0.28,0.07$ respectively. When he drinks tea, the probability that he has milk in it is 0.8 . When he drinks coffee, the probability that he has milk in it is 0.5 . When he drinks hot chocolate he always has milk in it.\\
(i) Draw a fully labelled tree diagram to represent this information.\\
(ii) Find the probability that Ayman's breakfast drink is coffee, given that his drink has milk in it.
\hfill \mbox{\textit{CAIE S1 2016 Q1 [5]}}