| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Combined probability with other distributions |
| Difficulty | Moderate -0.3 Part (i) is a straightforward binomial probability calculation requiring P(X ≥ 3) = 1 - P(X ≤ 2) with clearly stated parameters. Part (ii) involves conditional probability with a tree diagram but only spans two days with given probabilities, requiring basic complement rule application. Both parts are routine applications of standard techniques with no novel insight required, making this slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(P(3) + P(4) + P(5) = {}^5C_3\ 0.75^3 \times 0.25^2 +\) | M1 | One binomial term \({}^5C_x p^x (1-p)^{5-x}\), \(x \neq 0\) or 5, any \(p\) |
| \({}^5C_4\ 0.75^4 \times 0.25^1 + {}^5C_5\ 0.75^5 \times 0.25^0\) | M1 | Correct unsimplified expression |
| \(= 0.26367 + 0.39551 + 0.23730 = 0.896\ (459/512)\) | A1 | Correct final answer, allow 0.8965 (isw) but not 0.897 alone |
| Method 2: \(1 - P(0) - P(1) - P(2) = 1 - {}^5C_0\ 0.75^0 \times 0.25^5\) | M1 | One binomial term \({}^5C_x p^x (1-p)^{5-x}\), \(x \neq 0\) or 5, any \(p\) |
| \(-\ {}^5C_1\ 0.75^1 \times 0.25^4 -\ {}^5C_2\ 0.75^2 \times 0.25^3\) | M1 | Correct simplified expression |
| \(= 1 - 0.00097656 - 0.014648 - 0.087891 = 0.896\ (459/512)\) | A1 | Correct final answer, allow 0.8965 (isw) but not 0.897 alone |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(P(C,C) + P(C,C') + P(C',C)\); \(0.8 \times 0.9\) | B1 | Unsimplified prob completed on both days |
| \(0.8 \times 0.1 + 0.2 \times 0.6\) | M1 | Unsimplified prob \(0.8 \times a + 0.2 \times b\), \(a = 0.1\) or \(0.4\), \(b = 0.6\) or \(0.9\) |
| \(= 0.92\) oe | A1 | Correct final answer |
| Method 2: \(1 - P(C',C') = 1 - 0.2 \times 0.4\) | B1 | Unsimplified prob completed on no days |
| M1 | \(1 - 0.2 \times a\), \(a = 0.1\) or \(0.4\) allow unsimplified | |
| \(= 0.92\) | A1 | Correct final answer |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $P(3) + P(4) + P(5) = {}^5C_3\ 0.75^3 \times 0.25^2 +$ | **M1** | One binomial term ${}^5C_x p^x (1-p)^{5-x}$, $x \neq 0$ or 5, any $p$ |
| ${}^5C_4\ 0.75^4 \times 0.25^1 + {}^5C_5\ 0.75^5 \times 0.25^0$ | **M1** | Correct unsimplified expression |
| $= 0.26367 + 0.39551 + 0.23730 = 0.896\ (459/512)$ | **A1** | Correct final answer, allow 0.8965 (isw) but not 0.897 alone |
| **Method 2:** $1 - P(0) - P(1) - P(2) = 1 - {}^5C_0\ 0.75^0 \times 0.25^5$ | **M1** | One binomial term ${}^5C_x p^x (1-p)^{5-x}$, $x \neq 0$ or 5, any $p$ |
| $-\ {}^5C_1\ 0.75^1 \times 0.25^4 -\ {}^5C_2\ 0.75^2 \times 0.25^3$ | **M1** | Correct simplified expression |
| $= 1 - 0.00097656 - 0.014648 - 0.087891 = 0.896\ (459/512)$ | **A1** | Correct final answer, allow 0.8965 (isw) but not 0.897 alone |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $P(C,C) + P(C,C') + P(C',C)$; $0.8 \times 0.9$ | **B1** | Unsimplified prob completed on both days |
| $0.8 \times 0.1 + 0.2 \times 0.6$ | **M1** | Unsimplified prob $0.8 \times a + 0.2 \times b$, $a = 0.1$ or $0.4$, $b = 0.6$ or $0.9$ |
| $= 0.92$ oe | **A1** | Correct final answer |
| **Method 2:** $1 - P(C',C') = 1 - 0.2 \times 0.4$ | **B1** | Unsimplified prob completed on no days |
| | **M1** | $1 - 0.2 \times a$, $a = 0.1$ or $0.4$ allow unsimplified |
| $= 0.92$ | **A1** | Correct final answer |3 Jake attempts the crossword puzzle in his daily newspaper every day. The probability that he will complete the puzzle on any given day is 0.75 , independently of all other days.\\
(i) Find the probability that he will complete the puzzle at least three times over a period of five days.\\
Kenny also attempts the puzzle every day. The probability that he will complete the puzzle on a Monday is 0.8 . The probability that he will complete it on a Tuesday is 0.9 if he completed it on the previous day and 0.6 if he did not complete it on the previous day.\\
(ii) Find the probability that Kenny will complete the puzzle on at least one of the two days Monday and Tuesday in a randomly chosen week.\\
\hfill \mbox{\textit{CAIE S1 2018 Q3 [6]}}