| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.3 This is a standard multi-part combinations question covering basic selection (dividing groups), conditional probability, and arrangements with restrictions. All parts use routine techniques taught in S1: (a) uses C(9,3) with division by 2 correction, (b) is straightforward conditional probability, (c) uses complement counting, and (d) uses gap method. While multi-step, each part follows textbook methods without requiring novel insight, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^9C_6 \times {}^3C_3\) | M1 | \(^9C_k \times n\), \(k=6, 3\), \(n=1,2\); condone \(^9C_6 + {}^3C_3\), \(^9P_6 \times {}^3P_3\) |
| 84 | A1 | Accept unevaluated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Number with 3 Baker children \(= {}^6C_2\) or 15 | B1 | Correct seen anywhere, not multiplied or added |
| Total no. of selections \(= {}^9C_5\) or 126; \(\text{Probability} = \dfrac{\text{number of selections with 3 Baker children}}{\text{total number of selections}}\) | M1 | Seen as denominator of fraction |
| \(\dfrac{15}{126}\), \(0.119\) | A1 | OE, e.g. \(\dfrac{5}{42}\) |
| Alternative: \(\dfrac{3}{9}\times\dfrac{2}{8}\times\dfrac{1}{7}\left(\times\dfrac{6}{6}\right)\left(\times\dfrac{5}{5}\right)\times {}^5C_3\) | B1 | \({}^5C_3\) (OE) or 10 seen anywhere, multiplied by fractions only, not added |
| M1 | \(\dfrac{3}{9}\times\dfrac{2}{8}\times\dfrac{1}{7}\left(\times\dfrac{6}{6}\right)\left(\times\dfrac{5}{5}\right)\times k\), \(1\leqslant k\), \(k\) integer | |
| \(\dfrac{15}{126}\), \(0.119\) | A1 | OE, e.g. \(\dfrac{5}{42}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Total arrangements \(= 9!\)]; [Arrangements with men together \(= 8! \times 2\)]; Not together: \(9! - 8!\times 2\) | M1 | \(9! - k\) or \(362880 - k\), \(k\) an integer \(< 362880\) |
| \(8! \times 2\) | B1 | \(8!\times 2(!)\) or 80 640 seen anywhere |
| 282 240 | A1 | Exact value |
| Alternative: \(7!\times 8\times 7\) | B1 | \(7!\times k\), \(k\) positive integer \(> 1\) |
| M1 | \(m\times 8\times 7\), \(m\times {}^8P_2\), \(m\times {}^8C_2\), \(m\) positive integer \(> 1\) | |
| 282 240 | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(7!\times 2\times 7\) | M1 | \(7!\times k\), \(k\) positive integer \(> 1\); if \(7!\) not seen, condone \(7\times6\times5\times4\times3\times2\times(1)\times k\) or \(7\times6!\times k\) only |
| M1 | \(m\times 2\times 7\), \(m\) positive integer \(> 1\) | |
| 70 560 | A1 |
# Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^9C_6 \times {}^3C_3$ | M1 | $^9C_k \times n$, $k=6, 3$, $n=1,2$; condone $^9C_6 + {}^3C_3$, $^9P_6 \times {}^3P_3$ |
| 84 | A1 | Accept unevaluated |
# Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Number with 3 Baker children $= {}^6C_2$ or 15 | B1 | Correct seen anywhere, not multiplied or added |
| Total no. of selections $= {}^9C_5$ or 126; $\text{Probability} = \dfrac{\text{number of selections with 3 Baker children}}{\text{total number of selections}}$ | M1 | Seen as denominator of fraction |
| $\dfrac{15}{126}$, $0.119$ | A1 | OE, e.g. $\dfrac{5}{42}$ |
| **Alternative:** $\dfrac{3}{9}\times\dfrac{2}{8}\times\dfrac{1}{7}\left(\times\dfrac{6}{6}\right)\left(\times\dfrac{5}{5}\right)\times {}^5C_3$ | B1 | ${}^5C_3$ (OE) or 10 seen anywhere, multiplied by fractions only, not added |
| | M1 | $\dfrac{3}{9}\times\dfrac{2}{8}\times\dfrac{1}{7}\left(\times\dfrac{6}{6}\right)\left(\times\dfrac{5}{5}\right)\times k$, $1\leqslant k$, $k$ integer |
| $\dfrac{15}{126}$, $0.119$ | A1 | OE, e.g. $\dfrac{5}{42}$ |
# Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Total arrangements $= 9!$]; [Arrangements with men together $= 8! \times 2$]; Not together: $9! - 8!\times 2$ | M1 | $9! - k$ or $362880 - k$, $k$ an integer $< 362880$ |
| $8! \times 2$ | B1 | $8!\times 2(!)$ or 80 640 seen anywhere |
| 282 240 | A1 | Exact value |
| **Alternative:** $7!\times 8\times 7$ | B1 | $7!\times k$, $k$ positive integer $> 1$ |
| | M1 | $m\times 8\times 7$, $m\times {}^8P_2$, $m\times {}^8C_2$, $m$ positive integer $> 1$ |
| 282 240 | A1 | Exact value |
# Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $7!\times 2\times 7$ | M1 | $7!\times k$, $k$ positive integer $> 1$; if $7!$ not seen, condone $7\times6\times5\times4\times3\times2\times(1)\times k$ or $7\times6!\times k$ only |
| | M1 | $m\times 2\times 7$, $m$ positive integer $> 1$ |
| 70 560 | A1 | |6 Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children, are visiting an activity centre together. They will divide into groups for some of the activities.
\begin{enumerate}[label=(\alph*)]
\item In how many ways can the 9 people be divided into a group of 6 and a group of 3?\\
5 of the 9 people are selected at random for a particular activity.
\item Find the probability that this group of 5 people contains all 3 of the Baker children.\\
All 9 people stand in a line.
\item Find the number of different arrangements in which Mr Ahmed is not standing next to Mr Baker.
\item Find the number of different arrangements in which there is exactly one person between Mr Ahmed and Mr Baker.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q6 [11]}}