| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability with replacement/sequential selection |
| Difficulty | Moderate -0.8 Both parts are straightforward applications of basic counting principles: (i) requires recognizing 1 favorable outcome out of 6! arrangements, and (ii) involves a simple combination calculation C(6,3) with 1 favorable outcome. These are standard textbook exercises requiring only direct formula application with no problem-solving insight or multi-step reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(P(\text{all correct}) = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1} = \frac{1}{720}\) | M1 | For 6! Or 720 (sioc) or product of fractions |
| or \(P(\text{all correct}) = \frac{1}{6!} = \frac{1}{720} = 0.00139\) | A1 CAO | Accept 0.0014 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(P(\text{picks T, O, M}) = \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} = \frac{1}{20}\) | M1 | For denominators |
| or \(P(\text{picks T, O, M}) = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times 3! = \frac{1}{20}\) | M1 | For numerators or 3! |
| or \(P(\text{picks T, O, M}) = \frac{1}{\binom{6}{3}} = \frac{1}{20}\) | A1 CAO | Or M1 for \(\binom{6}{3}\) or 20 sioc; M1 for \(1/\binom{6}{3}\) |
| TOTAL | 5 |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $P(\text{all correct}) = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1} = \frac{1}{720}$ | M1 | For 6! Or 720 (sioc) or product of fractions |
| or $P(\text{all correct}) = \frac{1}{6!} = \frac{1}{720} = 0.00139$ | A1 CAO | Accept 0.0014 |
| **Total** | **2** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $P(\text{picks T, O, M}) = \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} = \frac{1}{20}$ | M1 | For denominators |
| or $P(\text{picks T, O, M}) = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times 3! = \frac{1}{20}$ | M1 | For numerators or 3! |
| or $P(\text{picks T, O, M}) = \frac{1}{\binom{6}{3}} = \frac{1}{20}$ | A1 CAO | Or M1 for $\binom{6}{3}$ or 20 sioc; M1 for $1/\binom{6}{3}$ |
| **TOTAL** | **5** | |
---2 Thomas has six tiles, each with a different letter of his name on it.\\
(i) Thomas arranges these letters in a random order. Find the probability that he arranges them in the correct order to spell his name.\\
(ii) On another occasion, Thomas picks three of the six letters at random. Find the probability that he picks the letters T, O and M (in any order).
\hfill \mbox{\textit{OCR MEI S1 2009 Q2 [5]}}