| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sum or difference of two spinners/dice |
| Difficulty | Moderate -0.3 This is a straightforward discrete probability distribution question requiring construction of a sample space (4×3=12 equally likely outcomes), calculating probabilities for X values, applying standard variance formula, and conditional probability. All techniques are routine S1 content with no novel insight required, making it slightly easier than average but not trivial due to the multi-part calculation work involved. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): \(-2,\ -1,\ 0,\ 1,\ 2,\ 3\); \(p\): \(\dfrac{1}{12},\ \dfrac{2}{12},\ \dfrac{3}{12},\ \dfrac{3}{12},\ \dfrac{2}{12},\ \dfrac{1}{12}\) | B1 | \(-2, -1, 0, 1, 2, 3\) seen as top line of a pdf table with at least 1 probability OR attempting to evaluate \(P(-2), P(-1), P(0), P(1), P(2), P(3)\) (condone additional values with \(p=0\) stated) |
| B1 | At least 4 probs correct (need not be in table) | |
| B1 | All probs correct in a table |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \dfrac{-2\times1 - 1\times2 + 0 + 1\times3 + 2\times2 + 1\times3}{12} = 0.5\) | M1 | Unsimplified expression for mean using *their* pdf table (or correct) with at least 2 non-zero values (may be seen in variance). Numerator terms may be implied by values. |
| \(\text{Var}(X) = \dfrac{(-2)^2\times1 + (-1)^2\times2 + 1^2\times3 + 2^2\times2 + 3^2\times1}{12} - (\textit{their}\ 0.5)^2\) | M1 | Unsimplified expression for variance using *their* pdf table (or correct) with at least 2 non-zero values and *their* \(E(X)\). Numerator terms may be implied by values. If \(-k^2\) is seen for \((-k)^2\), the method must be confirmed by seeing value used correctly |
| \(26/12 - 1/4 = 23/12\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \text{ non-zero}) = \frac{9}{12}\) | B1ft | If Binomial distribution used 0/3; \(P(X \text{ non-zero})\) ft from *their* pdf table, \(\Sigma p = 1\) oe |
| \(P(X=1 \mid X \text{ non-zero}) = \frac{P(X=1 \cap X \text{ non-zero})}{P(X \text{ non-zero})} = \frac{3/12}{9/12}\) | M1 | *Their* \(P(X=1)\)/their \(P(X \text{ non-zero})\) from *their* pdf table oe |
| \(= \frac{1}{3}\) oe | A1 | Correct final answer www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=1 \mid X \text{ non-zero}) = \frac{\text{Number of outcomes} = 1}{\text{Number of non-zero outcomes}}\) | B1ft | Number of non-zero outcomes (expect 9) ft from *their* outcome table or pdf table numerators oe |
| M1 | \(a/b\), \(a =\) *their* 3 from *their* outcome table or pdf table numerators, \(b =\) *their* 9 (not 12) | |
| \(= \frac{3}{9} = \frac{1}{3}\) oe | A1 | Correct final answer www |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: $-2,\ -1,\ 0,\ 1,\ 2,\ 3$; $p$: $\dfrac{1}{12},\ \dfrac{2}{12},\ \dfrac{3}{12},\ \dfrac{3}{12},\ \dfrac{2}{12},\ \dfrac{1}{12}$ | **B1** | $-2, -1, 0, 1, 2, 3$ seen as top line of a pdf table with at least 1 probability OR attempting to evaluate $P(-2), P(-1), P(0), P(1), P(2), P(3)$ (condone additional values with $p=0$ stated) |
| | **B1** | At least 4 probs correct (need not be in table) |
| | **B1** | All probs correct in a table |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \dfrac{-2\times1 - 1\times2 + 0 + 1\times3 + 2\times2 + 1\times3}{12} = 0.5$ | **M1** | Unsimplified expression for mean using *their* pdf table (or correct) with at least 2 non-zero values (may be seen in variance). Numerator terms may be implied by values. |
| $\text{Var}(X) = \dfrac{(-2)^2\times1 + (-1)^2\times2 + 1^2\times3 + 2^2\times2 + 3^2\times1}{12} - (\textit{their}\ 0.5)^2$ | **M1** | Unsimplified expression for variance using *their* pdf table (or correct) with at least 2 non-zero values and *their* $E(X)$. Numerator terms may be implied by values. If $-k^2$ is seen for $(-k)^2$, the method must be confirmed by seeing value used correctly |
| $26/12 - 1/4 = 23/12$ | **A1** | Correct final answer |
## Question 6(iii):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \text{ non-zero}) = \frac{9}{12}$ | B1ft | If Binomial distribution used 0/3; $P(X \text{ non-zero})$ ft from *their* pdf table, $\Sigma p = 1$ oe |
| $P(X=1 \mid X \text{ non-zero}) = \frac{P(X=1 \cap X \text{ non-zero})}{P(X \text{ non-zero})} = \frac{3/12}{9/12}$ | M1 | *Their* $P(X=1)$/their $P(X \text{ non-zero})$ from *their* pdf table oe |
| $= \frac{1}{3}$ oe | A1 | Correct final answer www |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=1 \mid X \text{ non-zero}) = \frac{\text{Number of outcomes} = 1}{\text{Number of non-zero outcomes}}$ | B1ft | Number of non-zero outcomes (expect 9) ft from *their* outcome table or pdf table numerators oe |
| | M1 | $a/b$, $a =$ *their* 3 from *their* outcome table or pdf table numerators, $b =$ *their* 9 (not 12) |
| $= \frac{3}{9} = \frac{1}{3}$ oe | A1 | Correct final answer www |
**Total: 3 marks**
---6 A fair red spinner has 4 sides, numbered 1,2,3,4. A fair blue spinner has 3 sides, numbered 1,2,3. When a spinner is spun, the score is the number on the side on which it lands. The spinners are spun at the same time. The random variable $X$ denotes the score on the red spinner minus the score on the blue spinner.\\
(i) Draw up the probability distribution table for $X$.\\
(ii) Find $\operatorname { Var } ( X )$.\\
(iii) Find the probability that $X$ is equal to 1 , given that $X$ is non-zero.\\
\hfill \mbox{\textit{CAIE S1 2018 Q6 [9]}}