| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Three or more independent values |
| Difficulty | Standard +0.3 This is a straightforward discrete probability question requiring basic probability calculations (finding probabilities from a ratio), standard expectation/variance formulas, and simple combinatorial reasoning for independent selections. The multi-part structure and need to consider different combinations elevates it slightly above average, but all techniques are routine for S1 level with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(3m) = 4/5\ (0.8)\), \(P(5m) = 1/5\ (0.2)\) | B1 | \(P(3m) = 4/5\) or \(P(5m) = 1/5\) seen or implied |
| \(E(X) = 17/5\ (3.4)\) | B1 | Correct \(E(X)\) |
| M1 | Subtract their mean\(^2\) numerically from \(\sum x^2 p\), no extra dividing | |
| \(\text{Var}(X) = 16/25\ (0.64)\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(3,5) + P(5,3) = 0.8 \times 0.2 + 0.2 \times 0.8\) | M1 | Summing two 2-factor terms |
| \(= 8/25\ (0.32)\) | A1\(\sqrt{}\) [2] | Correct answer, ft on \(2 \times p \times (1-p)\), their \(p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(11) = P(3,3,5) + P(3,5,3) + P(5,3,3)\) | M1 | Mult 2 probs for 3 with 1 prob for 5 |
| \(= (4/5 \times 4/5 \times 1/5) \times 3\) | M1 | Multiplying probs for 11 by 3 or summing 3 options |
| \(= 48/125\ (0.384)\) | A1 [3] | Correct final answer |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(3m) = 4/5\ (0.8)$, $P(5m) = 1/5\ (0.2)$ | B1 | $P(3m) = 4/5$ or $P(5m) = 1/5$ seen or implied |
| $E(X) = 17/5\ (3.4)$ | B1 | Correct $E(X)$ |
| | M1 | Subtract their mean$^2$ numerically from $\sum x^2 p$, no extra dividing |
| $\text{Var}(X) = 16/25\ (0.64)$ | A1 [4] | Correct answer |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(3,5) + P(5,3) = 0.8 \times 0.2 + 0.2 \times 0.8$ | M1 | Summing two 2-factor terms |
| $= 8/25\ (0.32)$ | A1$\sqrt{}$ [2] | Correct answer, ft on $2 \times p \times (1-p)$, their $p$ |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(11) = P(3,3,5) + P(3,5,3) + P(5,3,3)$ | M1 | Mult 2 probs for 3 with 1 prob for 5 |
| $= (4/5 \times 4/5 \times 1/5) \times 3$ | M1 | Multiplying probs for 11 by 3 or summing 3 options |
| $= 48/125\ (0.384)$ | A1 [3] | Correct final answer |
---3 A factory makes a large number of ropes with lengths either 3 m or 5 m . There are four times as many ropes of length 3 m as there are ropes of length 5 m .\\
(i) One rope is chosen at random. Find the expectation and variance of its length.\\
(ii) Two ropes are chosen at random. Find the probability that they have different lengths.\\
(iii) Three ropes are chosen at random. Find the probability that their total length is 11 m .
\hfill \mbox{\textit{CAIE S1 2011 Q3 [9]}}