| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection from categorized items |
| Difficulty | Moderate -0.8 This is a straightforward application of combinations with categorized items. Part (i) requires direct calculation of C(5,3) × C(9,3), while part (ii) adds one simple step of dividing by C(14,6). Both parts are standard textbook exercises requiring only recall of the combination formula and basic probability, with no problem-solving insight needed. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^9C_3 \times ^5C_3 = 84 \times 10 = 840\) | M1 | For either \(^9C_3\) or \(^5C_3\); zero for permutations |
| M1 | For product of both correct combinations | |
| A1 [3] | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total number of ways \(= ^{14}C_6 = 3003\) | M1 | For \(^{14}C_6\) seen in part (ii) |
| \(\text{Probability} = \frac{840}{3003} = \frac{40}{143} = 0.27972 = 0.280\) | M1 | For their \(840/3003\) or their \(840/^{14}C_6\) |
| A1 [3] | FT their 840; allow full marks for unsimplified fractional answers | |
| OR \(^6C_3 \times \frac{5}{14} \times \frac{4}{13} \times \frac{3}{12} \times \frac{9}{11} \times \frac{8}{10} \times \frac{7}{9} = 0.280\) | (M1) | For product of fractions; SC1 for \(^6C_3 \times (5/14)^3 \times (9/14)^3 = 0.2420\) |
| (M1)(A1) | For \(^6C_3\times\) correct product |
# Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^9C_3 \times ^5C_3 = 84 \times 10 = 840$ | M1 | For either $^9C_3$ or $^5C_3$; zero for permutations |
| | M1 | For product of both correct combinations |
| | A1 [3] | CAO |
# Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total number of ways $= ^{14}C_6 = 3003$ | M1 | For $^{14}C_6$ seen in part (ii) |
| $\text{Probability} = \frac{840}{3003} = \frac{40}{143} = 0.27972 = 0.280$ | M1 | For their $840/3003$ or their $840/^{14}C_6$ |
| | A1 [3] | FT their 840; allow full marks for unsimplified fractional answers |
| **OR** $^6C_3 \times \frac{5}{14} \times \frac{4}{13} \times \frac{3}{12} \times \frac{9}{11} \times \frac{8}{10} \times \frac{7}{9} = 0.280$ | (M1) | For product of fractions; SC1 for $^6C_3 \times (5/14)^3 \times (9/14)^3 = 0.2420$ |
| | (M1)(A1) | For $^6C_3\times$ correct product |2 An examination paper consists of two sections. Section A has 5 questions and Section B has 9 questions. Candidates are required to answer 6 questions.\\
(i) In how many different ways can a candidate choose 6 questions, if 3 are from Section A and 3 are from Section B?\\
(ii) Another candidate randomly chooses 6 questions to answer. Find the probability that this candidate chooses 3 questions from each section.
\hfill \mbox{\textit{OCR MEI S1 2012 Q2 [6]}}