| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multiple independent trials |
| Difficulty | Standard +0.3 This is a straightforward probability question with standard techniques: part (i) requires basic probability setup with constraints, part (ii) is a simple binomial calculation with ordering, and part (iii) is a routine normal approximation to binomial. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(P(2,4,6)\) all \(= p\), then \(P(1,3,5)\) all \(= 2p\) | 1 | M1 Using \(P(\text{even}) = 2P(\text{odd})\) or vice versa |
| \(3p + 6p = 1\) | 1 | M1 Summing \(P(\text{odd} + \text{even})\) or \(P(1,2,3,4,5,6) = 1\) |
| \(p = \frac{1}{9}\) so \(\text{prob}(3) = \frac{2}{9}\ (0.222)\) | 1 | A1 Correct answer |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(5,5,6) = \frac{2}{9} \times \frac{2}{9} \times \frac{1}{9} \times {}^3C_2\) | 2 | M1 Mult three probs together; M1 Mult by 3 or summing 3 options |
| \(= \frac{4}{243}\ (0.0165)\) | 1 | A1 Correct answer |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mu = 100 \times \frac{1}{3} = 33.3,\quad \sigma^2 = 100 \times \frac{1}{3} \times \frac{2}{3} = 22.2\) | 1 | B1 Unsimplified \(\frac{100}{3}\) and \(\frac{200}{9}\) seen |
| \(P(x \leq 37) = P\!\left(z \leq \dfrac{37.5 - \frac{100}{3}}{\sqrt{\frac{200}{9}}}\right) = P(z \leq 0.8839)\) | 3 | M1 Standardising, need sq rt; M1 36.5 or 37.5 seen; M1 correct area using their mean |
| \(= 0.812\) | 1 | A1 Correct answer |
| Total: 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $P(2,4,6)$ all $= p$, then $P(1,3,5)$ all $= 2p$ | 1 | **M1** Using $P(\text{even}) = 2P(\text{odd})$ or vice versa |
| $3p + 6p = 1$ | 1 | **M1** Summing $P(\text{odd} + \text{even})$ or $P(1,2,3,4,5,6) = 1$ |
| $p = \frac{1}{9}$ so $\text{prob}(3) = \frac{2}{9}\ (0.222)$ | 1 | **A1** Correct answer |
| **Total: 3** | | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(5,5,6) = \frac{2}{9} \times \frac{2}{9} \times \frac{1}{9} \times {}^3C_2$ | 2 | **M1** Mult three probs together; **M1** Mult by 3 or summing 3 options |
| $= \frac{4}{243}\ (0.0165)$ | 1 | **A1** Correct answer |
| **Total: 3** | | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu = 100 \times \frac{1}{3} = 33.3,\quad \sigma^2 = 100 \times \frac{1}{3} \times \frac{2}{3} = 22.2$ | 1 | **B1** Unsimplified $\frac{100}{3}$ and $\frac{200}{9}$ seen |
| $P(x \leq 37) = P\!\left(z \leq \dfrac{37.5 - \frac{100}{3}}{\sqrt{\frac{200}{9}}}\right) = P(z \leq 0.8839)$ | 3 | **M1** Standardising, need sq rt; **M1** 36.5 or 37.5 seen; **M1** correct area using their mean |
| $= 0.812$ | 1 | **A1** Correct answer |
| **Total: 5** | | |7 The faces of a biased die are numbered $1,2,3,4,5$ and 6 . The probabilities of throwing odd numbers are all the same. The probabilities of throwing even numbers are all the same. The probability of throwing an odd number is twice the probability of throwing an even number.\\
(i) Find the probability of throwing a 3 .\\
\includegraphics[max width=\textwidth, alt={}, center]{34ae4f06-d485-4138-82d8-902b70f08995-10_51_1563_495_331}\\
(ii) The die is thrown three times. Find the probability of throwing two 5 s and one 4 .\\
(iii) The die is thrown 100 times. Use an approximation to find the probability that an even number is thrown at most 37 times.\\
\hfill \mbox{\textit{CAIE S1 Q7 [11]}}