| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing standard formulas (P(7,5), combinations with restrictions) with minimal problem-solving required. Part (ii)(b) adds a simple conditional probability element but still follows routine methods. Easier than average A-level due to direct application of formulas without novel insight. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(^7P_5\) or \(\frac{7!}{2!}\) or \(7 \times 6 \times 5 \times 4 \times 3\) or \(^7C_5 \times 5!\) alone \(= 2520\) | M1, A1 | \(^7P_2\) or \(\frac{7!}{2!}\) M0A0; \(^7C_5 = 21\) or \(5! = 120\) M0A0 but see (i)(b) |
| Answer | Marks | Guidance |
|---|---|---|
| \(^6P_4\) or \(\frac{6!}{2!}\) or \(6 \times 5 \times 4 \times 3\) or \(^6C_4 \times 4!\) or \(360\) | M1 | alone or \(\times 2\) only; or \('2520' - 5 \times {^6P_4}\) M2 |
| \(\times 2\) | M1 | \(^6P_4 \times 2\) or \(6!\) alone M2; \(^6C_4 \times 2\) or \(6! \times 2\) alone M0M1 only; any other \(\times 2\) M0M0; or \('2520' \times \frac{2}{7}\) M2A0; (eg (ia)21 (ib) \(21 \times \frac{2}{7} = 6\) M2A0 but if ans is 6, must see wking); SC ONLY on ft from (i)(a): if (i)(a) \(5! = 120\), then (i)(b)\(4! \times 2 = 48\) alone M1M0A0; Other SC \(^5P_3 \times 2\) M2; NOT isw eg \(\frac{720}{2520} = \frac{2}{7}\) M1M1A0 |
| \(= 720\) | A1 cao |
| Answer | Marks |
|---|---|
| \(21\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(^5C_3\) or \(\frac{5!}{3!2!}\) or \(^5C_5\) seen or 10 seen in num | M1 | \(\frac{5}{7} \times \frac{4}{6}\) oe seen; Allow \(^5C_2\) seen BOD |
| \(\dfrac{^5C_3}{^5C_3 + ^5C_5}\) oe | M1 | \(\frac{5}{7} \times \frac{4}{6} \div \left(\frac{5}{7} \times \frac{4}{6} + \frac{2}{7} \times \frac{1}{6}\right)\) |
| \(\frac{10}{11}\) or \(0.909\) (3 sf) | A1 |
# Question 7(i)(a):
$^7P_5$ or $\frac{7!}{2!}$ or $7 \times 6 \times 5 \times 4 \times 3$ or $^7C_5 \times 5!$ alone $= 2520$ | M1, A1 | $^7P_2$ or $\frac{7!}{2!}$ M0A0; $^7C_5 = 21$ or $5! = 120$ M0A0 but see (i)(b)
**Total: [2]**
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# Question 7(i)(b):
$^6P_4$ or $\frac{6!}{2!}$ or $6 \times 5 \times 4 \times 3$ or $^6C_4 \times 4!$ or $360$ | M1 | alone or $\times 2$ only; or $'2520' - 5 \times {^6P_4}$ M2
$\times 2$ | M1 | $^6P_4 \times 2$ or $6!$ alone M2; $^6C_4 \times 2$ or $6! \times 2$ alone M0M1 only; any other $\times 2$ M0M0; or $'2520' \times \frac{2}{7}$ M2A0; (eg (ia)21 (ib) $21 \times \frac{2}{7} = 6$ M2A0 but if ans is 6, must see wking); SC ONLY on ft from (i)(a): if (i)(a) $5! = 120$, then (i)(b)$4! \times 2 = 48$ alone M1M0A0; Other SC $^5P_3 \times 2$ M2; NOT isw eg $\frac{720}{2520} = \frac{2}{7}$ M1M1A0
$= 720$ | A1 cao |
**Total: [3]**
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# Question 7(ii)(a):
$21$ | B1 |
**Total: [1]**
---
# Question 7(ii)(b):
$^5C_3$ or $\frac{5!}{3!2!}$ or $^5C_5$ seen or 10 seen in num | M1 | $\frac{5}{7} \times \frac{4}{6}$ oe seen; Allow $^5C_2$ seen BOD
$\dfrac{^5C_3}{^5C_3 + ^5C_5}$ oe | M1 | $\frac{5}{7} \times \frac{4}{6} \div \left(\frac{5}{7} \times \frac{4}{6} + \frac{2}{7} \times \frac{1}{6}\right)$
$\frac{10}{11}$ or $0.909$ (3 sf) | A1 |
**Total: [3]**
---7 (i) 5 of the 7 letters $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F } , \mathrm { G }$ are arranged in a random order in a straight line.
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements of 5 letters are possible?
\item How many of these arrangements end with a vowel (A or E)?\\
(ii) A group of 5 people is to be chosen from a list of 7 people.\\
(a) How many different groups of 5 people can be chosen?\\
(b) The list of 7 people includes Jill and Jo. A group of 5 people is chosen at random from the list. Given that either Jill and Jo are both chosen or neither of them is chosen, find the probability that both of them are chosen.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2012 Q7 [9]}}