| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Selection with type constraints |
| Difficulty | Moderate -0.8 This is a straightforward probability question requiring basic combinatorics (counting favorable outcomes over total outcomes) without replacement. The calculations involve simple combinations C(n,r) with small numbers, and the problem structure is standard for S1. It's easier than average because it's purely mechanical application of counting principles with no conceptual subtlety or multi-step reasoning required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6}{13}\times\frac{5}{13}\times\frac{3}{12} \times 3!\) oe | M1 | \(^6C_1 \times ^5C_1 \times ^3C_1 \div ^{13}C_3\) |
| M1 | With repl M0M1A0 | |
| \(= \frac{45}{182}\) or \(0.247\) (3 sfs)oe | A1 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6}{14}\times\frac{5}{13}\times\frac{3}{12} + \frac{6}{14}\times\frac{6}{13}\times\frac{3}{12} + \frac{6}{14}\times\frac{6}{13}\times\frac{4}{12}\) | M2 | \(^6C_1 + ^6C_1 + ^6C_1\) M1 for any one \((+ ^{12}C_3)\)M1 all 9 numerators correct, |
| \(= \frac{31}{364}\) or \(0.0852\) (3 sf) | A1 3 | With repl M1(6/14)³+(5/14)³+(4/14)³ |
### (i)
$\frac{6}{13}\times\frac{5}{13}\times\frac{3}{12} \times 3!$ oe | M1 | $^6C_1 \times ^5C_1 \times ^3C_1 \div ^{13}C_3$
| M1 | With repl M0M1A0
$= \frac{45}{182}$ or $0.247$ (3 sfs)oe | A1 3 |
### (ii)
$\frac{6}{14}\times\frac{5}{13}\times\frac{3}{12} + \frac{6}{14}\times\frac{6}{13}\times\frac{3}{12} + \frac{6}{14}\times\frac{6}{13}\times\frac{4}{12}$ | M2 | $^6C_1 + ^6C_1 + ^6C_1$ M1 for any one $(+ ^{12}C_3)$M1 all 9 numerators correct,
$= \frac{31}{364}$ or $0.0852$ (3 sf) | A1 3 | With repl M1(6/14)³+(5/14)³+(4/14)³
**Total [6]**
---A washing-up bowl contains 6 spoons, 5 forks and 3 knives. Three of these 14 items are removed at random, without replacement. Find the probability that
\begin{enumerate}[label=(\roman*)]
\item all three items are of different kinds, [3]
\item all three items are of the same kind. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q5 [6]}}