| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Both independence and mutual exclusivity |
| Difficulty | Standard +0.3 This question requires calculating probabilities from sample spaces and applying definitions of independence and mutual exclusivity. While it involves multiple steps (finding P(A), P(B), P(A∩B), then checking conditions), the concepts are straightforward and the calculations are routine for S1 level. The sample space for two dice is manageable, and both parts follow standard procedures without requiring novel insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}\) | M1 | Sensible attempt at \(P(A)\) |
| M1 | Sensible attempt at \(P(B)\) | |
| \(P(B) = \frac{27}{36} = \frac{3}{4}\) | B1 | Correct \(P(A \cap B)\) |
| \(P(A \cap B) = \frac{12}{36} = \frac{1}{3}\) | M1 | Cf \(P(A \cap B)\) with \(P(A) \times P(B)\); need at least 1 correct |
| \(P(A) \times P(B) = \frac{4}{9} \times \frac{3}{4} = \frac{1}{3}\) | ||
| Independent as \(P(A \cap B) = P(A) \times P(B)\) | A1 [5] | Correct conclusion following all correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Not mutually exclusive because \(P(A \cap B) \neq 0\); or give counter example e.g. 1 and 6 | B1\(\checkmark\) [1] | ft their \(P(A \cap B)\) |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}$ | M1 | Sensible attempt at $P(A)$ |
| | M1 | Sensible attempt at $P(B)$ |
| $P(B) = \frac{27}{36} = \frac{3}{4}$ | B1 | Correct $P(A \cap B)$ |
| $P(A \cap B) = \frac{12}{36} = \frac{1}{3}$ | M1 | Cf $P(A \cap B)$ with $P(A) \times P(B)$; need at least 1 correct |
| $P(A) \times P(B) = \frac{4}{9} \times \frac{3}{4} = \frac{1}{3}$ | | |
| Independent as $P(A \cap B) = P(A) \times P(B)$ | A1 [5] | Correct conclusion following all correct working |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Not mutually exclusive because $P(A \cap B) \neq 0$; or give counter example e.g. 1 and 6 | B1$\checkmark$ [1] | ft their $P(A \cap B)$ |
---3 Jason throws two fair dice, each with faces numbered 1 to 6 . Event $A$ is 'one of the numbers obtained is divisible by 3 and the other number is not divisible by 3 '. Event $B$ is 'the product of the two numbers obtained is even'.\\
(i) Determine whether events $A$ and $B$ are independent, showing your working.\\
(ii) Are events $A$ and $B$ mutually exclusive? Justify your answer.
\hfill \mbox{\textit{CAIE S1 2015 Q3 [6]}}