| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| 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 (independent events with replacement) and application of expectation formula E(X) = np. Both parts involve routine procedures with no problem-solving insight needed, making it easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Neither is an ace}) = \left(1 - \frac{4}{52}\right)^2 = \frac{2304}{2704} = \frac{144}{169} = 0.852\) | M1, A1 [2] | M1 for \(\frac{48}{52}\) or equivalent seen; CAO; Allow 0.85 with working |
| Answer | Marks | Guidance |
|---|---|---|
| Expected number \(= 10 \times 0.852 = 8.52\) | B1 [1] | FT their (i) if seen; Do not allow whole number final answer even if 8.52 seen first; Allow fractional answer |
## Question 2:
**(i)**
$P(\text{Neither is an ace}) = \left(1 - \frac{4}{52}\right)^2 = \frac{2304}{2704} = \frac{144}{169} = 0.852$ | M1, A1 [2] | M1 for $\frac{48}{52}$ or equivalent seen; CAO; Allow 0.85 with working
**(ii)**
Expected number $= 10 \times 0.852 = 8.52$ | B1 [1] | FT their (i) if seen; Do not allow whole number final answer even if 8.52 seen first; Allow fractional answer
---2 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 Q2 [3]}}