| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Bayes with complementary outcome |
| Difficulty | Moderate -0.8 This is a straightforward two-part Bayes' theorem question with clearly stated probabilities and standard tree diagram structure. Part (i) requires simple application of the law of total probability, and part (ii) is direct application of Bayes' theorem with complementary events. The numbers are clean and the context is simple, making this easier than average for A-level statistics. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{cup of coffee}) = 0.6\times0.9 + 0.4\times0.3 = 0.66\) | M1, A1 [2] | Summing two 2-factor probabilities; Correct answer, accept 0.660 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Not on time} \mid \text{no cup of coffee})\) | M1 | \(0.4\times0.7\) seen as numerator or denominator of a fraction |
| \(= \frac{P(\text{not on time} \cap \text{no cup})}{P(\text{no cup})} = \frac{0.4\times0.7}{1-0.66}\) | M1 | Attempt at \(P(\text{no cup})\) as \(0.1\times p_1 + 0.7\times p_2\) or as \(1-(i)\) seen anywhere |
| \(= \frac{0.28}{0.34} = 0.824\) | A1 [3] |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{cup of coffee}) = 0.6\times0.9 + 0.4\times0.3 = 0.66$ | M1, A1 [2] | Summing two 2-factor probabilities; Correct answer, accept 0.660 |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Not on time} \mid \text{no cup of coffee})$ | M1 | $0.4\times0.7$ seen as numerator or denominator of a fraction |
| $= \frac{P(\text{not on time} \cap \text{no cup})}{P(\text{no cup})} = \frac{0.4\times0.7}{1-0.66}$ | M1 | Attempt at $P(\text{no cup})$ as $0.1\times p_1 + 0.7\times p_2$ or as $1-(i)$ seen anywhere |
| $= \frac{0.28}{0.34} = 0.824$ | A1 [3] | |
---3 The probability that the school bus is on time on any particular day is 0.6 . If the bus is on time the probability that Sam the driver gets a cup of coffee is 0.9 . If the bus is not on time the probability that Sam gets a cup of coffee is 0.3 .\\
(i) Find the probability that Sam gets a cup of coffee.\\
(ii) Given that Sam does not get a cup of coffee, find the probability that the bus is not on time.
\hfill \mbox{\textit{CAIE S1 2016 Q3 [5]}}