| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Selection with replacement |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability with replacement using the multiplication and complement rules. Part (i) uses (19/20)^5, part (ii) applies binomial probability with n=5, p=1/20, and part (iii) requires recognizing the specific sequence (miss, miss, hit). All parts involve standard probability calculations with no conceptual subtlety or problem-solving insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((0.95)^5 = 0.774\) | M1 | For 0.95 seen, can be implied |
| A1 | For correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \((0.95)^4 \times (0.05)^1 \times {}_5C_1\) \(= 0.204\) | M1 | For any binomial calculation with 3 terms, powers summing to 5 |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \((0.95)^2 \times (0.05)\) \(= 0.0451(\frac{361}{8000})\) | M1 | For no Ps, no Cs, and only 3 terms of type \(p^r(1-p)\) |
| A1 | For correct answer |
**(i)** $(0.95)^5 = 0.774$ | M1 | For 0.95 seen, can be implied
A1 | For correct final answer
**2 marks total**
**(ii)** $(0.95)^4 \times (0.05)^1 \times {}_5C_1$ $= 0.204$ | M1 | For any binomial calculation with 3 terms, powers summing to 5
A1 | For correct answer
**2 marks total**
**(iii)** $(0.95)^2 \times (0.05)$ $= 0.0451(\frac{361}{8000})$ | M1 | For no Ps, no Cs, and only 3 terms of type $p^r(1-p)$
A1 | For correct answer
**2 marks total**
---4 Single cards, chosen at random, are given away with bars of chocolate. Each card shows a picture of one of 20 different football players. Richard needs just one picture to complete his collection. He buys 5 bars of chocolate and looks at all the pictures. Find the probability that\\
(i) Richard does not complete his collection,\\
(ii) he has the required picture exactly once,\\
(iii) he completes his collection with the third picture he looks at.
\hfill \mbox{\textit{CAIE S1 2003 Q4 [6]}}