| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring systematic enumeration of outcomes (5×3=15 equally likely cases), constructing a table, and applying standard formulas for mean, variance, and probability. The calculations are routine with no conceptual challenges beyond basic probability multiplication and expectation formulas, making it easier than average A-level material. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Score | 1 | 2 |
| Prob | \(\frac{3}{15}\) | \(\frac{4}{15}\) |
| Probability distribution table with correct scores | B1 | Allow extra score values if probability of zero stated |
| 2 probabilities (with correct score) correct | B1 | |
| 3 or more correct probabilities with correct scores | B1 | |
| \(\sum p = 1\), at least 4 probabilities | B1 | FT |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{mean} = \frac{(3+8+12+4+12+9)}{15} = \frac{48}{15}\ (3.2)\) | B1 | |
| \(\text{Var} = \frac{(3+16+36+16+72+81)}{15} - (\textit{their}\ 3.2)^2\) | M1 | FT substitute *their* attempts at scores in correct var formula, must have "– mean²" (condone probabilities not summing to 1) |
| \(= \frac{224}{15} - 3.2^2 = 4.69\ \left(\frac{352}{75}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Score of 4, 6, 9 | M1 | Identifying relevant scores from *their* mean and *their* table |
| \(\text{Prob}\ \frac{4}{15}\ (0.267)\) | A1 | Correct answer; SC B1 for 4/15 with no working |
## Question 6:
### Part 6(i):
| Score | 1 | 2 | 3 | 4 | 6 | 9 |
|-------|---|---|---|---|---|---|
| Prob | $\frac{3}{15}$ | $\frac{4}{15}$ | $\frac{4}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{15}$ |
Probability distribution table with correct scores | B1 | Allow extra score values if probability of zero stated
2 probabilities (with correct score) correct | B1 |
3 or more correct probabilities with correct scores | B1 |
$\sum p = 1$, at least 4 probabilities | B1 | FT
**Total: 4 marks**
### Part 6(ii):
$\text{mean} = \frac{(3+8+12+4+12+9)}{15} = \frac{48}{15}\ (3.2)$ | B1 |
$\text{Var} = \frac{(3+16+36+16+72+81)}{15} - (\textit{their}\ 3.2)^2$ | M1 | FT substitute *their* attempts at scores in correct var formula, must have "– mean²" (condone probabilities not summing to 1)
$= \frac{224}{15} - 3.2^2 = 4.69\ \left(\frac{352}{75}\right)$ | A1 |
**Total: 3 marks**
### Part 6(iii):
Score of 4, 6, 9 | M1 | Identifying relevant scores from *their* mean and *their* table
$\text{Prob}\ \frac{4}{15}\ (0.267)$ | A1 | Correct answer; SC B1 for 4/15 with no working
**Total: 2 marks**
---6 A fair five-sided spinner has sides numbered 1, 1, 1, 2, 3. A fair three-sided spinner has sides numbered $1,2,3$. Both spinners are spun once and the score is the product of the numbers on the sides the spinners land on.\\
(i) Draw up the probability distribution table for the score.\\
\includegraphics[max width=\textwidth, alt={}, center]{da4a61b9-f55d-40ed-a721-a6aee962f0d6-08_67_1569_484_328}\\
(ii) Find the mean and the variance of the score.\\
(iii) Find the probability that the score is greater than the mean score.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [9]}}