| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability distributions from selection |
| Difficulty | Standard +0.3 This is a straightforward application of combinations to calculate probabilities without replacement. Part (i) requires basic counting principles, part (ii) is a standard 'show that' calculation, and part (iii) involves computing a complete probability distribution using the same technique repeatedly. While it requires careful organization across multiple cases, the methods are routine for S1 level with no novel problem-solving insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{all different}) = \dfrac{{}_{3}C_1 \times {}_{4}C_1 \times {}_{5}C_1}{{}_{12}C_3}\) | M1 | Attempt using combinations, with \({}_{12}C_3\) denom, or \(P(RGY)\) in any order, i.e. \(12 \times 11 \times 10\) in denom |
| M1 | Correct numerator, or multiplying by 6 | |
| \(= 3/11\ (= 0.273)\) | A1 3 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{exactly } 2\ G) = \dfrac{{}_{4}C_2 \times {}_{8}C_1}{{}_{12}C_3}\) | M1 | Attempt using combinations, or mult any \(P(GG\bar{G}) \times 3\) or \(P(GGY) \times 3 + P(GGR) \times 3\) |
| \(= 12/55\ AG\) | A1 2 | Correct answer AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\): 0, 1, 2, 3; \(P(X=x)\): \(14/55\), \(28/55\), \(12/55\), \(1/55\); decimal: 0.255, 0.509, 0.218, 0.018 | M1 | For seeing \(P(0, 1, 2, 3)\) only and 1 or more probs |
| M1 | For reasonable attempt at \(P(X = 0\) or \(1\) or \(3)\) | |
| A1 | For one correct probability seen other than \(P(X=2)\) | |
| A1 | For a second probability correct other than \(P(X=2)\) | |
| A1 5 | All correct |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{all different}) = \dfrac{{}_{3}C_1 \times {}_{4}C_1 \times {}_{5}C_1}{{}_{12}C_3}$ | M1 | Attempt using combinations, with ${}_{12}C_3$ denom, or $P(RGY)$ in any order, i.e. $12 \times 11 \times 10$ in denom |
| | M1 | Correct numerator, or multiplying by 6 |
| $= 3/11\ (= 0.273)$ | A1 **3** | Correct answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{exactly } 2\ G) = \dfrac{{}_{4}C_2 \times {}_{8}C_1}{{}_{12}C_3}$ | M1 | Attempt using combinations, or mult any $P(GG\bar{G}) \times 3$ or $P(GGY) \times 3 + P(GGR) \times 3$ |
| $= 12/55\ AG$ | A1 **2** | Correct answer AG |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$: 0, 1, 2, 3; $P(X=x)$: $14/55$, $28/55$, $12/55$, $1/55$; decimal: 0.255, 0.509, 0.218, 0.018 | M1 | For seeing $P(0, 1, 2, 3)$ only and 1 or more probs |
| | M1 | For reasonable attempt at $P(X = 0$ or $1$ or $3)$ |
| | A1 | For one correct probability seen other than $P(X=2)$ |
| | A1 | For a second probability correct other than $P(X=2)$ |
| | A1 **5** | All correct |7 A vegetable basket contains 12 peppers, of which 3 are red, 4 are green and 5 are yellow. Three peppers are taken, at random and without replacement, from the basket.\\
(i) Find the probability that the three peppers are all different colours.\\
(ii) Show that the probability that exactly 2 of the peppers taken are green is $\frac { 12 } { 55 }$.\\
(iii) The number of green peppers taken is denoted by the discrete random variable $X$. Draw up a probability distribution table for $X$.
\hfill \mbox{\textit{CAIE S1 2007 Q7 [10]}}