| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Sum or combination of independent binomial values |
| Difficulty | Standard +0.3 Part (i) is a straightforward binomial probability calculation (P(X ≥ 2) where X ~ B(4, 1/3)), requiring only complement rule or direct summation. Part (ii) requires systematic enumeration of ways to partition 5 into four parts from {1,2,3}, which adds mild combinatorial reasoning but remains routine for S1 level. Both parts use standard techniques with no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p = 1/3\); \(P(\geq 2) = 1 - P(0,1) = 1 - (2/3)^4 - {}^4C_1(1/3)(2/3)^3\) | M1 | Bin term \({}^4C_x p^x(1-p)^{4-x}\), \(0 < p < 1\) |
| or \(P(2,3,4) = {}^4C_2(1/3)^2(2/3)^2 + {}^4C_3(1/3)^3(2/3) + (1/3)^4\) | M1 | Correct unsimplified answer |
| \(= \dfrac{11}{27}\), \(0.407\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{sum is } 5) = P(1,1,1,2) \times 4 = (1/3)^4 \times 4\) | M1 | 1, 1, 1, 2 seen or 4 options |
| M1 | Multiply by \((1/3)^4\) | |
| \(= \dfrac{4}{81}\), \(0.0494\) | A1 [3] |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p = 1/3$; $P(\geq 2) = 1 - P(0,1) = 1 - (2/3)^4 - {}^4C_1(1/3)(2/3)^3$ | M1 | Bin term ${}^4C_x p^x(1-p)^{4-x}$, $0 < p < 1$ |
| or $P(2,3,4) = {}^4C_2(1/3)^2(2/3)^2 + {}^4C_3(1/3)^3(2/3) + (1/3)^4$ | M1 | Correct unsimplified answer |
| $= \dfrac{11}{27}$, $0.407$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{sum is } 5) = P(1,1,1,2) \times 4 = (1/3)^4 \times 4$ | M1 | 1, 1, 1, 2 seen or 4 options |
| | M1 | Multiply by $(1/3)^4$ |
| $= \dfrac{4}{81}$, $0.0494$ | A1 [3] | |
---2 A fair triangular spinner has three sides numbered 1, 2, 3. When the spinner is spun, the score is the number of the side on which it lands. The spinner is spun four times.\\
(i) Find the probability that at least two of the scores are 3 .\\
(ii) Find the probability that the sum of the four scores is 5 .
\hfill \mbox{\textit{CAIE S1 2016 Q2 [6]}}