| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Code/password formation |
| Difficulty | Easy -1.2 This is a straightforward probability question requiring basic counting principles. Part (i) is trivial (1/10000), and part (ii) only requires knowing there are 4! = 24 arrangements of 4 distinct digits, giving probability 1/24. No complex reasoning or problem-solving is needed—just direct application of basic permutation formulas. |
| Spec | 2.03a Mutually exclusive and independent events |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{guess correctly}) = 0.1^4 = 0.0001\) | B1 CAO | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{guess correctly}) = \frac{1}{4!} = \frac{1}{24}\) | M1; A1 CAO | 2 marks |
## Question 5:
### Part (i)
$P(\text{guess correctly}) = 0.1^4 = 0.0001$ | B1 CAO | **1 mark**
### Part (ii)
$P(\text{guess correctly}) = \frac{1}{4!} = \frac{1}{24}$ | M1; A1 CAO | **2 marks**
---5 My credit card has a 4-digit code called a PIN. You should assume that any 4-digit number from 0000 to 9999 can be a PIN.\\
(i) If I cannot remember any digits and guess my number, find the probability that I guess it correctly.
In fact my PIN consists of four different digits. I can remember all four digits, but cannot remember the correct order.\\
(ii) If I now guess my number, find the probability that I guess it correctly.
\hfill \mbox{\textit{OCR MEI S1 Q5 [3]}}