| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Repeated trials with selection |
| Difficulty | Standard +0.3 This is a straightforward application of combinations followed by binomial probability. Part (i) requires calculating C(5,2)×C(5,1)/C(10,3), which is standard selection probability. Part (ii) applies the binomial distribution using the result from (i), requiring calculation of P(X≥3) for n=4 trials. Both parts are routine S1 techniques with no conceptual challenges beyond careful arithmetic. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | M1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1FT | M1FT | M1FT |
## Part (i)
**Answer:**
$$3 \times \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8} = \frac{300}{720} = \frac{5}{12} = (0.4167)$$
Or:
$$\left(\frac{5}{2}\right) \times \left(\frac{5}{1}\right) \div \left(\frac{10}{3}\right) = \frac{10 \times 5}{120} = \frac{5}{12}$$
**Marks:** M1 | M1 | M1 | A1 | [4]
**Guidance:**
- M1: For 5/10 × 4/9 or Correct working but then multiplied or divided by some factor scores M1M1M0A0
- M1: For × 5/8
- M1: For 3 × triple product
- A1: CAO (Fully simplified). Answer must be a fraction
- Alternative marks available: M1* for $\left(\frac{5}{2}\right) \times \left(\frac{5}{1}\right)$ Seen; M1* for $\left(\frac{10}{3}\right)$ Seen; M1* for whole fraction Correct working but then multiplied or divided by some factor scores M1M1M0A0
## Part (ii)
**Answer:**
$$4 \times \frac{7}{12} \times \left(\frac{5}{12}\right)^3 + \left(\frac{5}{12}\right)^3$$
$$= 0.169 + 0.030 = 0.199$$
Or:
$$\frac{875}{5184} + \frac{625}{20736} = \frac{1375}{6912}$$
**Marks:** M1FT | M1FT | M1FT | A1 | [4]
**Guidance:**
- M1FT: For first probability. Allow ³C₃
- M1FT: For (5/12)³
- M1FT: For sum of both correct probabilities. Provided sum < 1
- A1: CAO. Do not allow 0.2, unless fuller answer seen first. Alternative for 1- (P(0)+P(1)+P(2)) allow M1FT for two 'correct' probs, M1 for sum of three 'correct', M1 for 1 – answer, A1 CAO
---2 Every evening, 5 men and 5 women are chosen to take part in a phone-in competition. Of these 10 people, exactly 3 will win a prize. These 3 prize-winners are chosen at random.\\
(i) Find the probability that, on a particular evening, 2 of the prize-winners are women and the other is a man. Give your answer as a fraction in its simplest form.\\
(ii) Four evenings are selected at random. Find the probability that, on at least three of the four evenings, 2 of the prize-winners are women and the other is a man.
\hfill \mbox{\textit{OCR MEI S1 2013 Q2 [8]}}