| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward probability distribution question requiring only basic recall and routine calculations: (i) uses the sum-to-1 constraint to find p algebraically, (ii) applies standard E(X) and Var(X) formulas, and (iii) uses conditional probability P(X=2|X>0). All parts are textbook exercises with no problem-solving insight needed, making this easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | - 1 | 0 | 1 | 2 | 4 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 4 }\) | \(p\) | \(p\) | \(\frac { 3 } { 8 }\) | \(4 p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{4} + p + p + \frac{3}{8} + 4p = 1\) | M1 | Unsimplified sum of probabilities equated to 1 |
| \(p = \frac{1}{16}\) | A1 | If method FT from *their* incorrect (i), expressions for \(E(X)\) and \(\text{Var}(X)\) must be seen unsimplified with all probabilities \(<1\), condone not adding to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = -\frac{1}{4} + \frac{1}{16} + \frac{6}{8} + 1 = \frac{25}{16}\) | M1 | May be implied by use in Variance, accept unsimplified |
| \(\text{Var}(X) = \frac{1}{4} + \frac{1}{16} + \frac{12}{8} + \frac{16}{4} - \left(\frac{25}{16}\right)^2\) | M1 | Substitute into correct variance formula, must have '− their mean²' |
| \(\frac{863}{256}\) or \(3.37\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=2 \mid X>0) = \frac{P(X=2)}{P(X>0)} = \frac{\frac{3}{8}}{\frac{11}{16}}\) | M1 | Conditional probability formula used consistent with their probabilities |
| \(\frac{6}{11}\) or \(0.545\) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{4} + p + p + \frac{3}{8} + 4p = 1$ | M1 | Unsimplified sum of probabilities equated to 1 |
| $p = \frac{1}{16}$ | A1 | If method FT from *their* incorrect (i), expressions for $E(X)$ and $\text{Var}(X)$ must be seen unsimplified with all probabilities $<1$, condone not adding to 1 |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = -\frac{1}{4} + \frac{1}{16} + \frac{6}{8} + 1 = \frac{25}{16}$ | M1 | May be implied by use in Variance, accept unsimplified |
| $\text{Var}(X) = \frac{1}{4} + \frac{1}{16} + \frac{12}{8} + \frac{16}{4} - \left(\frac{25}{16}\right)^2$ | M1 | Substitute into correct variance formula, must have '− their mean²' |
| $\frac{863}{256}$ or $3.37$ | A1 | OE |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=2 \mid X>0) = \frac{P(X=2)}{P(X>0)} = \frac{\frac{3}{8}}{\frac{11}{16}}$ | M1 | Conditional probability formula used consistent with their probabilities |
| $\frac{6}{11}$ or $0.545$ | A1 | |
---4 In a probability distribution the random variable $X$ takes the values $- 1,0,1,2,4$. The probability distribution table for $X$ is as follows.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 1 & 0 & 1 & 2 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 4 }$ & $p$ & $p$ & $\frac { 3 } { 8 }$ & $4 p$ \\
\hline
\end{tabular}
\end{center}
(i) Find the value of $p$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
(iii) Given that $X$ is greater than zero, find the probability that $X$ is equal to 2 .\\
\hfill \mbox{\textit{CAIE S1 2019 Q4 [7]}}