| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multi-item selection from population |
| Difficulty | Easy -1.3 This is a straightforward S1 probability question testing basic combinations and counting principles. All three parts involve standard 'without replacement' calculations with small numbers that can be solved by simple enumeration or basic combination formulas. Part (i) is P(2 vowels from 5 cards) = C(2,2)/C(5,2) = 1/10, part (ii) uses C(2,1)×C(3,2)/C(5,3), and part (iii) is simply 4/5. No problem-solving insight required—purely mechanical application of taught methods to a textbook scenario. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space5.01a Permutations and combinations: evaluate probabilities |
### (i)
**Answer:**
- $\frac{1}{5}\times \frac{1}{4} \times 2$ or $\frac{2}{5}\times \frac{1}{4}$ alone oe
- $= \frac{1}{10}$ or 0.1 oe
**Marks:** M1, A1[2]
**Guidance:**
- Allow M1 for $\frac{1}{5} \times \frac{1}{4}$, but NOT other methods leading to $\frac{1}{20}$ and NOT $\frac{1}{20}$ with no wking
- M1 for totally correct method except $\frac{1}{5} \times \frac{1}{4}$ seen: M1
- NB $\frac{2}{5}\times \frac{1}{4}\times 2$ M0A0; $\frac{2}{5}\times \frac{1}{4}$ M0A0
### (ii)
**Answer:**
- $\frac{2}{5}\times \frac{1}{4}\times \frac{2}{3}$ or $\frac{1}{5}\times \frac{1}{4}\times \frac{2}{3} \times 2$ oe or correct list of 6 comb's with 1 vowel or $\frac{2}{5}\times \frac{1}{3}\times \frac{2}{3}$ or $\frac{6}{20}$ or $\frac{3}{10}$ or similar or $\frac{1}{5}$ or 0.2 oe
**Marks:** M1, A1[3]
**Guidance:**
- Only if using complement (ie 1–P(0V or 2V)):
- $\frac{3}{5}\times \frac{2}{5}\times \frac{1}{3}$ or $\frac{2}{5}\times \frac{3}{5}\times \frac{1}{3}$ M1
- or $1-(\frac{3}{5}\times \frac{2}{5}\times \frac{1}{3} + \frac{2}{5}\times \frac{3}{5}\times \frac{1}{3})$ M1
- $1 - (\frac{3}{5}\times \frac{2}{5}\times \frac{1}{3} + \frac{2}{5}\times \frac{3}{5}\times \frac{1}{3} \times 3)$ M1
- $= \frac{2}{5}$ or 0.6 oe B1
### (iii)
**Answer:**
- $1-\frac{1}{5} \times \frac{4}{4}\times \frac{3}{2}$ or $1-\frac{1}{5}$ or $\frac{1-\frac{1}{5}}{1}$ or $\frac{1×^4C_3}{5×^4C_4}$ or $\frac{1}{5}×4$ = $\frac{4}{5}$ or 0.8 oe
**Marks:** M1, A1[2]
**Guidance:**
- $\frac{4}{5} \times \ldots$ M0A0, eg $\frac{4}{5} \times \frac{1}{5}$ M0A0
- See comment before 6(i)
---The diagram shows five cards, each with a letter on it.
\includegraphics{figure_6}
The letters A and E are vowels; the letters B, C and D are consonants.
\begin{enumerate}[label=(\roman*)]
\item Two of the five cards are chosen at random, without replacement. Find the probability that they both have vowels on them. [2]
\item The two cards are replaced. Now three of the five cards are chosen at random, without replacement. Find the probability that they include exactly one card with a vowel on it. [3]
\item The three cards are replaced. Now four of the five cards are chosen at random without replacement. Find the probability that they include the card with the letter B on it. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q6 [7]}}