| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability distributions from selection |
| Difficulty | Standard +0.3 This is a straightforward hypergeometric distribution problem requiring counting letters (12 total, 5 vowels), calculating combinations for P(V=0,1,2,3), and applying standard expectation/variance formulas. The given answer in part (a) provides scaffolding, making it slightly easier than average but still requiring systematic probability calculation across multiple cases. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks |
|---|---|
| \(P(V = 1) = 3 \times \frac{5}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{21}{44}\) | M2 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(V = 3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{1}{22}\) | M2 A2 | |
| \(V\) | 0 | 1 |
| \(P(V = v)\) | \(\frac{7}{44}\) | \(\frac{21}{44}\) |
| \(E(V) = \sum v P(v) = 0 + \frac{21}{44} + \frac{14}{22} + \frac{3}{22} = \frac{4}{4}\) | M1 A1 | |
| \(E(V^2) = \sum v^2 P(v) = 0 + \frac{21}{44} + \frac{28}{22} + \frac{3}{22} = \frac{95}{44}\) | M1 A1 | |
| \(\text{Var}(V) = \frac{95}{44} - (\frac{4}{4})^2 = \frac{105}{176}\) or \(0.597\) (3sf) | M1 A1 | (13) |
5 vowels, 7 consonants
$P(V = 1) = 3 \times \frac{5}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{21}{44}$ | M2 A1 |
$P(V = 0) = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{7}{44}$
$P(V = 2) = 3 \times \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} = \frac{7}{22}$
$P(V = 3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{1}{22}$ | M2 A2 |
| $V$ | 0 | 1 | 2 | 3 |
|-----|-------|--------|--------|--------|
| $P(V = v)$ | $\frac{7}{44}$ | $\frac{21}{44}$ | $\frac{7}{22}$ | $\frac{1}{22}$ |
$E(V) = \sum v P(v) = 0 + \frac{21}{44} + \frac{14}{22} + \frac{3}{22} = \frac{4}{4}$ | M1 A1 |
$E(V^2) = \sum v^2 P(v) = 0 + \frac{21}{44} + \frac{28}{22} + \frac{3}{22} = \frac{95}{44}$ | M1 A1 |
$\text{Var}(V) = \frac{95}{44} - (\frac{4}{4})^2 = \frac{105}{176}$ or $0.597$ (3sf) | M1 A1 | (13)5. The letters of the word DISTRIBUTION are written on separate cards. The cards are then shuffled and the top three are turned over.
Let the random variable $V$ be the number of vowels that are turned over.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( V = 1 ) = \frac { 21 } { 44 }$.
\item Find the probability distribution of $V$.
\item Find $\mathrm { E } ( V )$ and $\operatorname { Var } ( V )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [13]}}