| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.8 This is a straightforward two-equation, two-unknown problem using basic probability axioms (sum = 1) and expectation formula. The algebra is simple and the variance calculation in part (b) is routine application of Var(X) = E(X²) - [E(X)]². No conceptual difficulty or problem-solving insight required. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.4 | \(p\) | \(r\) | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p + r + 0.55 = 1\) | M1 | Using sum of probabilities \(= 1\) to form an equation. Accept \(p + r = 0.45\) oe. |
| \(p + 2r + 0.45 = 1.1\) | M1 | Use \(E(X) = 1.1\) to form an equation. Accept \(p + 2r = 0.65\) oe. NB: These marks can be gained in either order; the second M may have an algebraic substitution. |
| \(p = 0.25,\ r = 0.2\) | A1 | If both Ms not awarded, SC B1 for \(p = 0.25,\ r = 0.2\) stated. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{Var}(X) =]\ \left[0.4 \times 0^2 + \textit{their}\ 0.25 \times 1^2\right] + (\textit{their}\ 0.2) \times 2^2 + 0.15 \times 3^2 - 1.1^2\) \([= [0+]\ 0.25 + 0.8 + 1.35 - 1.21]\) | M1 | Correct formula for variance method using their probability distribution table, \(0 <\) their \(P(x) < 1\). |
| \(= 1.19,\ 1\dfrac{19}{100}\) | A1 | If M0 awarded, SC B1 for \(1.19\) www. \(\dfrac{119}{100}\) is A0. |
| 2 |
## Question 1(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $p + r + 0.55 = 1$ | M1 | Using sum of probabilities $= 1$ to form an equation. Accept $p + r = 0.45$ oe. |
| $p + 2r + 0.45 = 1.1$ | M1 | Use $E(X) = 1.1$ to form an equation. Accept $p + 2r = 0.65$ oe. NB: These marks can be gained in either order; the second M may have an algebraic substitution. |
| $p = 0.25,\ r = 0.2$ | A1 | If both Ms not awarded, **SC B1** for $p = 0.25,\ r = 0.2$ stated. |
| | **3** | |
---
## Question 1(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Var}(X) =]\ \left[0.4 \times 0^2 + \textit{their}\ 0.25 \times 1^2\right] + (\textit{their}\ 0.2) \times 2^2 + 0.15 \times 3^2 - 1.1^2$ $[= [0+]\ 0.25 + 0.8 + 1.35 - 1.21]$ | M1 | Correct formula for variance method using their probability distribution table, $0 <$ their $P(x) < 1$. |
| $= 1.19,\ 1\dfrac{19}{100}$ | A1 | If M0 awarded, **SC B1** for $1.19$ www. $\dfrac{119}{100}$ is A0. |
| | **2** | |1 A competitor in a throwing event has three attempts to throw a ball as far as possible. The random variable $X$ denotes the number of throws that exceed 30 metres. The probability distribution table for $X$ is shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.4 & $p$ & $r$ & 0.15 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { E } ( X ) = 1.1$, find the value of $p$ and the value of $r$.
\item Find the numerical value of $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q1 [5]}}