| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Verify conditions in context |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question requiring recall of conditions and standard probability calculations. Part (i) is pure recall, part (ii) is a routine cumulative probability calculation with given parameters, and part (iii) applies binomial distribution twice in a standard nested manner. No novel problem-solving or insight required—easier than average A-level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Constant probability (of completing) | B1 | Any one condition of these two |
| Independent trials/events | B1 | The other condition |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(5,6,7) = {}^7C_5(0.7)^5(0.3)^2 + {}^7C_6(0.7)^6(0.3)^1 + (0.7)^7\) | M1 | Bin term \({}^7C_x(0.7)^x(0.3)^{7-x}\), \(x \neq 0, 7\) |
| A1 | Correct unsimplified answer (sum) OE | |
| \(= 0.647\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(0,1,2,3,4) = 1 - \text{their } 0.6471 = 0.3529\) | M1 | Find \(P(\leqslant 4)\) either by subtracting their (ii) from 1, or from adding probs of \(0,1,2,3,4\) with \(n=7\) (or 10) and \(p = 0.7\) |
| \(P(3) = {}^{10}C_3(0.3529)^3(0.6471)^7\) | M1 | \({}^{10}C_3(\text{their } 0.353)^3(1 - \text{their } 0.353)^7\) on its own |
| \(= 0.251\) | A1 | |
| Total: 3 |
# Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Constant probability (of completing) | B1 | Any one condition of these two |
| Independent trials/events | B1 | The other condition |
| **Total: 2** | | |
# Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(5,6,7) = {}^7C_5(0.7)^5(0.3)^2 + {}^7C_6(0.7)^6(0.3)^1 + (0.7)^7$ | M1 | Bin term ${}^7C_x(0.7)^x(0.3)^{7-x}$, $x \neq 0, 7$ |
| | A1 | Correct unsimplified answer (sum) OE |
| $= 0.647$ | A1 | |
| **Total: 3** | | |
# Question 5(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(0,1,2,3,4) = 1 - \text{their } 0.6471 = 0.3529$ | M1 | Find $P(\leqslant 4)$ either by subtracting their (ii) from 1, or from adding probs of $0,1,2,3,4$ with $n=7$ (or 10) and $p = 0.7$ |
| $P(3) = {}^{10}C_3(0.3529)^3(0.6471)^7$ | M1 | ${}^{10}C_3(\text{their } 0.353)^3(1 - \text{their } 0.353)^7$ on its own |
| $= 0.251$ | A1 | |
| **Total: 3** | | |5 Hebe attempts a crossword puzzle every day. The number of puzzles she completes in a week (7 days) is denoted by $X$.\\
(i) State two conditions that are required for $X$ to have a binomial distribution.\\
On average, Hebe completes 7 out of 10 of these puzzles.\\
(ii) Use a binomial distribution to find the probability that Hebe completes at least 5 puzzles in a week.\\
(iii) Use a binomial distribution to find the probability that, over the next 10 weeks, Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks.\\
\hfill \mbox{\textit{CAIE S1 2017 Q5 [8]}}