| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probabilities in table form with k |
| Difficulty | Moderate -0.8 This is a straightforward application of the fundamental probability axiom that probabilities sum to 1. Students simply need to add the expressions (4p + 5p² + 1.5p + 2.5p + 1.5p = 1), collect like terms to get 5p² + 10p - 1 = 0, and solve the quadratic. It requires only basic algebraic manipulation with no conceptual difficulty or problem-solving insight. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general |
| \(x\) | 1 | 4 | 5 | 7 | 9 |
| \(\mathrm { P } ( X = x )\) | \(4 p\) | \(5 p ^ { 2 }\) | \(1.5 p\) | \(2.5 p\) | \(1.5 p\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1\) and \(10p^2 + 19p - 2 = 0\) | M1 | Summing 5 probs to = 1 can be implied |
| \(p = 0.1\) or \(-2\) | A1 | For 0.1 seen with or without –2 |
| \(p = 0.1\) | A1 [3] | Choosing 0.1 must be by rejecting –2 |
$4p + 5p^2 + 1.5p + 2.5p + 1.5p = 1$ and $10p^2 + 19p - 2 = 0$ | M1 | Summing 5 probs to = 1 can be implied
$p = 0.1$ or $-2$ | A1 | For 0.1 seen with or without –2
$p = 0.1$ | A1 [3] | Choosing 0.1 must be by rejecting –21 The discrete random variable $X$ takes the values 1, 4, 5, 7 and 9 only. The probability distribution of $X$ is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 4 & 5 & 7 & 9 \\
\hline
$\mathrm { P } ( X = x )$ & $4 p$ & $5 p ^ { 2 }$ & $1.5 p$ & $2.5 p$ & $1.5 p$ \\
\hline
\end{tabular}
\end{center}
Find $p$.
\hfill \mbox{\textit{CAIE S1 2010 Q1 [3]}}