| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Single batch expected count |
| Difficulty | Easy -1.2 This is a straightforward binomial distribution question requiring only basic probability calculation (48/52 × 48/52) and then multiplying by n=10 for expected value. It involves simple arithmetic with no conceptual challenges beyond recognizing independence and applying E(X)=np, making it easier than average A-level material. |
| Spec | 2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Neither is an ace}) = \left(1 - \frac{4}{52}\right)^2\) | M1 | For 48/52 oe seen |
| \(= \frac{2304}{2704} = \frac{144}{169} = 0.852\) | A1 | CAO. Allow 0.85 with working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected number \(= 10 \times 0.852 = 8.52\) | B1 | FT their (i) if seen. Do not allow whole number final answer even if 8.52 seen first. Allow fractional answer |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Neither is an ace}) = \left(1 - \frac{4}{52}\right)^2$ | M1 | For 48/52 oe seen |
| $= \frac{2304}{2704} = \frac{144}{169} = 0.852$ | A1 | CAO. Allow 0.85 with working |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected number $= 10 \times 0.852 = 8.52$ | B1 | FT their (i) if seen. Do not allow whole number final answer even if 8.52 seen first. Allow fractional answer |
---3 A normal pack of 52 playing cards contains 4 aces. A card is drawn at random from the pack. It is then replaced and the pack is shuffled, after which another card is drawn at random.\\
(i) Find the probability that neither card is an ace.\\
(ii) This process is repeated 10 times. Find the expected number of times for which neither card is an ace.
\hfill \mbox{\textit{OCR MEI S1 2015 Q3 [3]}}