| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.8 This is a straightforward application of combinations with standard probability calculation. Part (i) is direct C(25,15), part (ii) requires multiplying C(13,8)×C(12,7), and part (iii) divides these results. All steps are routine with no conceptual challenges beyond recognizing the basic combination structure. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\binom{25}{15} = 3268760\) | M1 | Accept \(^{25}C_{15}\) or \(\frac{25!}{(15!10!)}\) or equivalent. No marks for permutations |
| A1 | Exact answer required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\binom{13}{8} \times \binom{12}{7} = 1287 \times 792 = 1019304\) | M1 | For product of both correct combinations. No marks for permutations |
| A1 | CAO. Exact answer required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1019304}{3268760} = 0.312\) | M1 | For their (ii) divided by their (i) |
| Allow fully simplified fraction \(\frac{11583}{37145}\) | A1 FT | Allow 0.31 with working |
| OR: \(\binom{15}{8} \times \frac{13}{25} \times \frac{12}{24} \times \cdots \times \frac{7}{18} \times \frac{12}{17} \times \frac{11}{16} \times \cdots \times \frac{6}{11}\) | (M1) | For product of fractions with coefficient. SC1 for \(\binom{15}{8} \times \left(\frac{13}{25}\right)^8 \times \left(\frac{12}{25}\right)^7\) |
| \(= 0.312\) | (A1) |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\binom{25}{15} = 3268760$ | M1 | Accept $^{25}C_{15}$ or $\frac{25!}{(15!10!)}$ or equivalent. No marks for permutations |
| | A1 | Exact answer required |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\binom{13}{8} \times \binom{12}{7} = 1287 \times 792 = 1019304$ | M1 | For product of both correct combinations. No marks for permutations |
| | A1 | CAO. Exact answer required |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1019304}{3268760} = 0.312$ | M1 | For their (ii) divided by their (i) |
| Allow fully simplified fraction $\frac{11583}{37145}$ | A1 FT | Allow 0.31 with working |
| **OR:** $\binom{15}{8} \times \frac{13}{25} \times \frac{12}{24} \times \cdots \times \frac{7}{18} \times \frac{12}{17} \times \frac{11}{16} \times \cdots \times \frac{6}{11}$ | (M1) | For product of fractions with coefficient. SC1 for $\binom{15}{8} \times \left(\frac{13}{25}\right)^8 \times \left(\frac{12}{25}\right)^7$ |
| $= 0.312$ | (A1) | |
---4 A rugby team of 15 people is to be selected from a squad of 25 players.\\
(i) How many different teams are possible?\\
(ii) In fact the team has to consist of 8 forwards and 7 backs. If 13 of the squad are forwards and the other 12 are backs, how many different teams are now possible?\\
(iii) Find the probability that, if the team is selected at random from the squad of 25 players, it contains the correct numbers of forwards and backs.
\hfill \mbox{\textit{OCR MEI S1 2015 Q4 [6]}}