| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.8 This is a straightforward combinatorics question testing basic counting principles. Part (i) requires recognizing permutations with repetition (7!/2!). Part (ii) involves simple counting of combinations with a repeated element, requiring only careful enumeration rather than complex reasoning. All parts are standard textbook exercises with no novel problem-solving required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(7!\) or 5040 or \(^7P_7\) seen | M1 | or \(5!\times(^6C_2 + 6)\) NOT \(5!\times^6C_2\); or \(\frac{2}{7}\times\frac{1}{6}\times\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}\times\frac{1}{2}\) alone M2; or \(\geq 5\) correct fracs mult; or 6 correct fracs mult \(\times\ldots\) M1 |
| \(1 \div \frac{7!}{2}\) or \(\frac{2}{7!}\) | M1 | \(\frac{1}{5!\times(6C2+6)}\) |
| \(= \frac{1}{2520}\) or \(0.000397\) (3 sf) | A1 | or \(\frac{2}{5040}\) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | B1 | Ignore any working seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(^5C_2\) alone (or \(\times\ ^2C_2\)) or \(^6C_3\div2(!)\) or \(\frac{2}{7}\times^7C_3\) or \(^5P_2\div2\) | M1 | alone, eg NOT \(^5C_2\times\ldots\) or \(^5C_2+\ldots\); But allow \(^5C_2\) as denom of prob M1A0 |
| \(= 10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 + 10 + ^5C_3\) | M1 | or \(^6C_3 +\) "5" or \(^7C_3 -\) "10" or \(^7C_3 - ^5C_2\); ft (a) &/or (b) only if working seen; Allow as denom of a prob M1A0 |
| \(= 25\) | A1f |
# Question 6:
## Part (i)
$7!$ or 5040 or $^7P_7$ seen | M1 | or $5!\times(^6C_2 + 6)$ NOT $5!\times^6C_2$; or $\frac{2}{7}\times\frac{1}{6}\times\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}\times\frac{1}{2}$ alone M2; or $\geq 5$ correct fracs mult; or 6 correct fracs mult $\times\ldots$ M1
$1 \div \frac{7!}{2}$ or $\frac{2}{7!}$ | M1 | $\frac{1}{5!\times(6C2+6)}$
$= \frac{1}{2520}$ or $0.000397$ (3 sf) | A1 | or $\frac{2}{5040}$
**[3]**
## Part (ii)(a)
5 | B1 | Ignore any working seen
**[1]**
## Part (ii)(b)
$^5C_2$ alone (or $\times\ ^2C_2$) or $^6C_3\div2(!)$ or $\frac{2}{7}\times^7C_3$ or $^5P_2\div2$ | M1 | alone, eg NOT $^5C_2\times\ldots$ or $^5C_2+\ldots$; But allow $^5C_2$ as denom of prob M1A0
$= 10$ | A1 |
**[2]**
## Part (ii)(c)
$5 + 10 + ^5C_3$ | M1 | or $^6C_3 +$ "5" or $^7C_3 -$ "10" or $^7C_3 - ^5C_2$; ft (a) &/or (b) only if working seen; Allow as denom of a prob M1A0
$= 25$ | A1f |
**[2]**6 (i) The seven digits $1,1,2,3,4,5,6$ are arranged in a random order in a line. Find the probability that they form the number 1452163.\\
(ii) Three of the seven digits $1,1,2,3,4,5,6$ are chosen at random, without regard to order.
\begin{enumerate}[label=(\alph*)]
\item How many possible groups of three digits contain two 1s?
\item How many possible groups of three digits contain exactly one 1?
\item How many possible groups of three digits can be formed altogether?
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2015 Q6 [8]}}