| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.3 This is a straightforward combinations question requiring basic probability calculations using C(n,r). Part (i) involves simple complementary counting, and part (ii) requires summing a few cases with at-most-2-girls complement. Both are standard S1 techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Total number of selections \(= {}^{12}C_7 = 792\) | B1 | Seen as denominator of fraction |
| Selections with boy included \(= {}^{11}C_6\) or \({}^{12}C_7 - {}^{11}C_7 = 462\) | M1 | Correct unsimplified expression for selections with boy included seen as numerator of fraction |
| Probability \(= 462/792 = 7/12\ (0.583)\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| prob of boy not included \(= \frac{11}{12} \times \frac{10}{11} \times \ldots \times \frac{5}{6} = 5/12\) | B1 | Correct unsimplified prob |
| \(1 - 5/12\) | M1 | Subtracting prob from 1 |
| \(= 7/12\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(2G + 5B\): \({}^4C_2 \times {}^8C_5 = 336\) | B1 | One unsimplified product correct |
| \(3G + 4B\): \({}^4C_3 \times {}^8C_4 = 280\); \(4G + 3B\): \({}^4C_4 \times {}^8C_3 = 56\) | M1 | No of selections added for 2, 3 and 4 girls with no of boys summing to 7 |
| Total \(= 672\) | A1 | Correct total |
| Probability \(= 672/792\ (28/33)\ (0.848)\) | A1ft | Correct answer – 'total'/('total no of selections' from i) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0G + 7B\): \({}^4C_0 \times {}^8C_7 = 8\) | B1 | One unsimplified no of selections correct |
| \(1G + 6B\): \({}^4C_1 \times {}^8C_6 = 112\); Total \(= 120\) | M1 | No of selections added for 0 and 1 girls with no of boys summing to 7 |
| \(({}^{12}C_7 - 120)/792\) or \(1 - 120/792\) | A1 | \(792 - 120 = 672\) or \(1 - 120/792\) |
| Probability \(= 672/792\ (28/33)\ (0.848)\) | A1ft | '672' over '792' from i |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - P(0) - P(1) = 1 - \left(\frac{8}{12} \times \frac{7}{11} \times \ldots \times \frac{2}{6}\right) - \left(\frac{8}{12} \times \ldots \times \frac{3}{7} \times \frac{4}{6} \times 7\right)\) | B1 | One correct unsimplified prob for 0 or 1 |
| \(= 1 - 1/99 - 14/99\) | M1 | Subtracting 'P(0)' and 'P(1)' using products of 7 fractions with denominators from 12 to 6, from 1 |
| A1 | Both probs correct unsimplified | |
| \(= 84/99 = 28/33\) | A1ft | \(1 - \text{'P(0)'} - \text{'P(1)'}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(2) + P(3) + P(4) = 42/99 + 35/99 + 7/99\) | B1 | One correct unsimplified prob for 2, 3 or 4 |
| M1 | Adding 'P(2)', 'P(3)' and P(4)' using products of 7 fractions with denominators from 12 to 6 | |
| A1 | Three probs correct unsimplified | |
| \(= 84/99 = 28/33\) | A1ft | \(\text{'P(2)'} + \text{'P(3)'} + \text{'P(4)'}\) |
## Question 4(i):
Total number of selections $= {}^{12}C_7 = 792$ | B1 | Seen as denominator of fraction
Selections with boy included $= {}^{11}C_6$ or ${}^{12}C_7 - {}^{11}C_7 = 462$ | M1 | Correct unsimplified expression for selections with boy included seen as numerator of fraction
Probability $= 462/792 = 7/12\ (0.583)$ | A1 | Correct answer
OR
prob of boy not included $= \frac{11}{12} \times \frac{10}{11} \times \ldots \times \frac{5}{6} = 5/12$ | B1 | Correct unsimplified prob
$1 - 5/12$ | M1 | Subtracting prob from 1
$= 7/12$ | A1 | Correct answer
---
## Question 4(ii):
**Method 1:**
$2G + 5B$: ${}^4C_2 \times {}^8C_5 = 336$ | B1 | One unsimplified product correct
$3G + 4B$: ${}^4C_3 \times {}^8C_4 = 280$; $4G + 3B$: ${}^4C_4 \times {}^8C_3 = 56$ | M1 | No of selections added for 2, 3 and 4 girls with no of boys summing to 7
Total $= 672$ | A1 | Correct total
Probability $= 672/792\ (28/33)\ (0.848)$ | A1ft | Correct answer – 'total'/('total no of selections' from i)
**Method 2:**
$0G + 7B$: ${}^4C_0 \times {}^8C_7 = 8$ | B1 | One unsimplified no of selections correct
$1G + 6B$: ${}^4C_1 \times {}^8C_6 = 112$; Total $= 120$ | M1 | No of selections added for 0 and 1 girls with no of boys summing to 7
$({}^{12}C_7 - 120)/792$ or $1 - 120/792$ | A1 | $792 - 120 = 672$ or $1 - 120/792$
Probability $= 672/792\ (28/33)\ (0.848)$ | A1ft | '672' over '792' from i
**Method 3 (probability):**
$1 - P(0) - P(1) = 1 - \left(\frac{8}{12} \times \frac{7}{11} \times \ldots \times \frac{2}{6}\right) - \left(\frac{8}{12} \times \ldots \times \frac{3}{7} \times \frac{4}{6} \times 7\right)$ | B1 | One correct unsimplified prob for 0 or 1
$= 1 - 1/99 - 14/99$ | M1 | Subtracting 'P(0)' and 'P(1)' using products of 7 fractions with denominators from 12 to 6, from 1
| A1 | Both probs correct unsimplified
$= 84/99 = 28/33$ | A1ft | $1 - \text{'P(0)'} - \text{'P(1)'}$
**Method 4 (probability):**
$P(2) + P(3) + P(4) = 42/99 + 35/99 + 7/99$ | B1 | One correct unsimplified prob for 2, 3 or 4
| M1 | Adding 'P(2)', 'P(3)' and P(4)' using products of 7 fractions with denominators from 12 to 6
| A1 | Three probs correct unsimplified
$= 84/99 = 28/33$ | A1ft | $\text{'P(2)'} + \text{'P(3)'} + \text{'P(4)'}$
---4 Out of a class of 8 boys and 4 girls, a group of 7 people is chosen at random.\\
(i) Find the probability that the group of 7 includes one particular boy.\\
(ii) Find the probability that the group of 7 includes at least 2 girls.\\
\hfill \mbox{\textit{CAIE S1 2018 Q4 [7]}}