| 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 | Moderate -0.3 Part (i) is a straightforward probability without replacement calculation requiring basic counting principles. Part (ii) involves setting up and solving a simple linear equation from given probability information. Both parts use standard S1 techniques with no conceptual challenges, making this slightly easier than average but not trivial due to the multi-step nature and algebraic manipulation required in part (ii). |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
### (i)
**Answer:**
- $12\times 10\times 5$ (in numerators or alone) OR any prod of 3 probs×6(or ×3! or ×P₃)
- $\frac{12}{27}\times \frac{10}{26}\times \frac{5}{25}\times 6$ or $\frac{12×10×5}{27×C_3}$
- $= \frac{8}{39}$ oe or 0.205 (3 sfs)
**Marks:** M1, M1, A1[3]
**Guidance:**
- or $^{12}C_1\times ^{10}C_1\times ^5C_1$ or 600 (in numerators or alone)
- or eg $(\frac{12}{27}\times \frac{10}{26}\times \frac{5}{25}+\frac{12}{27}\times \frac{5}{25}\times \frac{10}{26})$ ×3 Fully correct method
- Examples:
- $\frac{12}{27}\times \frac{10}{26}\times \frac{5}{25} \times 6$ or $\frac{12}{27}\times \frac{10}{26}\times \frac{5}{25}$ ×3 M1M0A0
- or $\frac{1}{27}\times \frac{1}{26}\times \frac{1}{25} \times 6$ M1M0A0
### (ii)
**Answer:**
- $0.4 \times \frac{x}{50}$ OR $0.6 \times \frac{50-x}{50}$ oe or $0.4\times \frac{x}{50} + 0.6 \times \frac{50-x}{50} = 0.54$
- $4x = 60$ oe, two terms A1
- $p = 0.3$ or $b = 0.7$ A1
- T & I: $0.4\times \frac{x}{50}$ or etc OR one trial ($n \neq 15$) M1
- Trial of $n = 15$ M1A1
- Answer stated A1
**Marks:** M1, A1, A1, [4]
**Guidance:**
- $0.4 \times p$ OR $0.6 \times (1-p)$ or similar M1
- $0.4\times p + 0.6\times (1-p) = 0.54$ M1
- If $x\to(50-x)$ or $p\to (1-p)$: similar mks including 1st A1 for $p = 0.7$ or $x = 35$ M1
- Correct answer scores full marks unless clearly from incorrect method.
---\begin{enumerate}[label=(\roman*)]
\item A bag contains 12 black discs, 10 white discs and 5 green discs. Three discs are drawn at random from the bag, without replacement. Find the probability that all three discs are of different colours. [3]
\item A bag contains 30 red discs and 20 blue discs. A second bag contains 50 discs, each of which is either red or blue. A disc is drawn at random from each bag. The probability that these two discs are of different colours is 0.54. Find the number of red discs that were in the second bag at the start. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q8 [7]}}