| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection from categorized items |
| Difficulty | Moderate -0.8 This is a straightforward combinations question requiring only basic counting principles (multiplication rule) and simple probability calculations. Part (i) is routine multiplication (4×5×3), part (ii) involves counting ordered pairs with restrictions (also straightforward), and part (iii) applies basic probability concepts. No novel problem-solving or complex reasoning required—below average difficulty for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks |
|---|---|
| \(4 \times 5 \times 3 = 60\) | B1 M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Number of ways two people choose different starters from 4: \(\binom{4}{2} = 6\) | B1 | Shown |
| Answer | Marks |
|---|---|
| Different starters: 6, different mains: \(\binom{5}{2}=10\), different sweets: \(\binom{3}{2}=3\) | M1 |
| \(6 \times 10 \times 3 = 180\) | A1 |
| Answer | Marks |
|---|---|
| \(P(\text{same main}) = \frac{1}{5}\) | B1 |
| Answer | Marks |
|---|---|
| \(P(\text{different for every course}) = \frac{3}{4} \times \frac{4}{5} \times \frac{2}{3}\) | M1 |
| \(= \frac{24}{60} = \frac{2}{5}\) | A1 |
# Question 4:
## Part (i)
| $4 \times 5 \times 3 = 60$ | B1 M1 A1 | |
## Part (ii)(A)
| Number of ways two people choose different starters from 4: $\binom{4}{2} = 6$ | B1 | Shown |
## Part (ii)(B)
| Different starters: 6, different mains: $\binom{5}{2}=10$, different sweets: $\binom{3}{2}=3$ | M1 | |
| $6 \times 10 \times 3 = 180$ | A1 | |
## Part (iii)(A)
| $P(\text{same main}) = \frac{1}{5}$ | B1 | |
## Part (iii)(B)
| $P(\text{different for every course}) = \frac{3}{4} \times \frac{4}{5} \times \frac{2}{3}$ | M1 | |
| $= \frac{24}{60} = \frac{2}{5}$ | A1 | |
---4 Peter and Esther visit a restaurant for a three-course meal. On the menu there are 4 starters, 5 main courses and 3 sweets. Peter and Esther each order a starter, a main course and a sweet.
\begin{enumerate}[label=(\roman*)]
\item Calculate the number of ways in which Peter may choose his three-course meal.
\item Suppose that Peter and Esther choose different dishes from each other.\\
(A) Show that the number of possible combinations of starters is 6 .\\
(B) Calculate the number of possible combinations of 6 dishes for both meals.
\item Suppose instead that Peter and Esther choose their dishes independently.\\
(A) Write down the probability that they choose the same main course.\\
(B) Find the probability that they choose different dishes from each other for every course.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2006 Q4 [7]}}