| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.3 This is a standard discrete probability distribution question requiring systematic application of probability axioms (sum to 1) and expectation formulas. While it involves solving simultaneous equations and computing variance, the steps are routine and well-practiced in S1. The algebraic manipulation is straightforward, making it slightly easier than average for A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X=4) = 3P(X=2)]\) \(4k(4+a) = 3 \times 2k(2+a)\) \(16k + 4ak = 12k + 6ak\) | M1 | Using \(P(X=4) = 3P(X=2)\) to form an equation in \(a\) and \(k\) |
| \(a = 2\) | A1 | If M0 scored, SC B1 for \(a=2\) www |
| \(3k + 8k + 15k + 24k = 1\) | M1 | Using sum of probabilities \(= 1\) to form an equation in \(k\): \(k(1+a) + 2k(2+a) + 3k(3+a) + 4k(4+a) = 1\) |
| \(k = \frac{1}{50}\) | A1 | If M0 scored, SC B1 for \(k = \frac{1}{50}\) www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X\): 1, 2, 3, 4; \(P(X)\): \(\frac{3}{50}, 0.06\); \(\frac{8}{50}, 0.16\); \(\frac{15}{50}, 0.3\); \(\frac{24}{50}, 0.48\) | B1 FT | \(P(X)\): \(k(1+a)\), \(2k(2+a)\), \(3k(3+a)\), \(4k(4+a)\). \(0 < p < 1\) for all outcomes, must be numerical |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(X) = \frac{3}{50}\times1 + \frac{8}{50}\times2^2 + \frac{15}{50}\times3^2 + \frac{24}{50}\times4^2 - 3.2^2\) | M1 | Correct formula for variance method from their probability distribution table, \(0 \leq \text{their } P(x) \leq 1\). Accept \(\frac{3+32+135+384}{50} - \frac{256}{25}\) |
| \([= 11.08 - 3.2^2 =] \ 0.84[0], \ \frac{21}{25}\) | A1 | If M0 score SC B1 for 0.84 www |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X=4) = 3P(X=2)]$ $4k(4+a) = 3 \times 2k(2+a)$ $16k + 4ak = 12k + 6ak$ | M1 | Using $P(X=4) = 3P(X=2)$ to form an equation in $a$ and $k$ |
| $a = 2$ | A1 | If M0 scored, SC B1 for $a=2$ www |
| $3k + 8k + 15k + 24k = 1$ | M1 | Using sum of probabilities $= 1$ to form an equation in $k$: $k(1+a) + 2k(2+a) + 3k(3+a) + 4k(4+a) = 1$ |
| $k = \frac{1}{50}$ | A1 | If M0 scored, SC B1 for $k = \frac{1}{50}$ www |
---
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X$: 1, 2, 3, 4; $P(X)$: $\frac{3}{50}, 0.06$; $\frac{8}{50}, 0.16$; $\frac{15}{50}, 0.3$; $\frac{24}{50}, 0.48$ | B1 FT | $P(X)$: $k(1+a)$, $2k(2+a)$, $3k(3+a)$, $4k(4+a)$. $0 < p < 1$ for all outcomes, must be numerical |
---
## Question 3(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(X) = \frac{3}{50}\times1 + \frac{8}{50}\times2^2 + \frac{15}{50}\times3^2 + \frac{24}{50}\times4^2 - 3.2^2$ | M1 | Correct formula for variance method from their probability distribution table, $0 \leq \text{their } P(x) \leq 1$. Accept $\frac{3+32+135+384}{50} - \frac{256}{25}$ |
| $[= 11.08 - 3.2^2 =] \ 0.84[0], \ \frac{21}{25}$ | A1 | If M0 score SC B1 for 0.84 www |
---3 The random variable $X$ takes the values $1,2,3,4$. It is given that $\mathrm { P } ( X = x ) = k x ( x + a )$, where $k$ and $a$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { P } ( X = 4 ) = 3 \mathrm { P } ( X = 2 )$, find the value of $a$ and the value of $k$.
\item Draw up the probability distribution table for $X$, giving the probabilities as numerical fractions.
\item Given that $\mathrm { E } ( X ) = 3.2$, find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q3 [7]}}