| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | November |
| Marks | 4 |
| 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 S1 question requiring only two standard equations (probabilities sum to 1, and expectation formula) to solve a simple simultaneous system. The algebra is routine with no conceptual challenges, making it easier than average A-level maths questions. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| \(x\) | 1 | 3 | 5 | 7 |
| \(\mathrm { P } ( X = x )\) | 0.3 | \(a\) | \(b\) | 0.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i): \(a + b = 0.45\) | B1 | Accept unsimplified equation |
| Part (ii): \(0.3 + 3a + 5b + 7 \times 0.25 = 4\) | M1 | For an equation involving \(\sum x_i p_i = 4\) must be correct unsimplified version, seen anywhere |
| M1 | For sensible attempt to solve the two equations ie eliminating one letter | |
| A1 | For correct \(a\) and \(b\) | |
| \(a = 0.15, \quad b = 0.3\) | A1 | 3 |
**Part (i):** $a + b = 0.45$ | B1 | Accept unsimplified equation
**Part (ii):** $0.3 + 3a + 5b + 7 \times 0.25 = 4$ | M1 | For an equation involving $\sum x_i p_i = 4$ must be correct unsimplified version, seen anywhere
| M1 | For sensible attempt to solve the two equations ie eliminating one letter
| A1 | For correct $a$ and $b$
$a = 0.15, \quad b = 0.3$ | A1 | 3
---1 The discrete random variable $X$ has the following probability distribution.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
$x$ & 1 & 3 & 5 & 7 \\
\hline
$\mathrm { P } ( X = x )$ & 0.3 & $a$ & $b$ & 0.25 \\
\hline
\end{tabular}
\end{center}
(i) Write down an equation satisfied by $a$ and $b$.\\
(ii) Given that $\mathrm { E } ( X ) = 4$, find $a$ and $b$.
\hfill \mbox{\textit{CAIE S1 2002 Q1 [4]}}