| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | At least one success |
| Difficulty | Moderate -0.3 Part (i) is a straightforward binomial probability calculation using P(X ≤ 7) = 1 - P(X > 7), requiring only calculator work. Part (ii) involves solving 1 - (0.65)^n > 0.99 for n using logarithms, which is a standard 'at least one success' problem with routine algebraic manipulation. Both parts are typical S1 exercises with no novel insight required, making this slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{at most } 7) = 1 - P(8,9,10) = 1 - {}^{10}C_8(0.35)^8(0.65)^2 - {}^{10}C_9(0.35)^9(0.65)^1 - (0.35)^{10}\) | M1 | Use of normal approximation M0. Binomial term of form \({}^{10}C_x p^x(1-p)^{10-x}\), \(0 |
| \([= 1 - 0.004281 - 0.0005123 - 0.00002759]\) | A1 | Correct unsimplified (or individual terms evaluated) answer seen. Condone \(1-A+B+C\) leading to correct solution |
| \(= 0.995\) | B1 | B1 not dependent on previous marks |
| Alternative: \(P(\text{at most }7) = P(0,1,2,3,4,5,6,7)\) | M1 | Binomial term of form \({}^{10}C_x p^x(1-p)^{10-x}\), \(0 |
| \(= (0.65)^{10} + {}^{10}C_1(0.35)^1(0.65)^9 + \ldots + {}^{10}C_7(0.35)^7(0.65)^3\) | A1 | Correct unsimplified answer or individual terms evaluated seen |
| \(= 0.995\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1-(0.65)^n > 0.99\), \(\ 0.01 > (0.65)^n\) | M1 | Equation or inequality with \((0.65)^n\) and \(0.01\) or \((0.35)^n\) and \(0.99\) only. (Note \(1-0.99\) is equivalent to \(0.01\) etc.) |
| \(n > 10.69\) | M1 | Solving \(a^n = c\), \(0 |
| smallest \(n = 11\) | A1 | CAO |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{at most } 7) = 1 - P(8,9,10) = 1 - {}^{10}C_8(0.35)^8(0.65)^2 - {}^{10}C_9(0.35)^9(0.65)^1 - (0.35)^{10}$ | M1 | Use of normal approximation M0. Binomial term of form ${}^{10}C_x p^x(1-p)^{10-x}$, $0<p<1$ any $p$, $x\neq 10,0$ |
| $[= 1 - 0.004281 - 0.0005123 - 0.00002759]$ | A1 | Correct unsimplified (or individual terms evaluated) answer seen. Condone $1-A+B+C$ leading to correct solution |
| $= 0.995$ | B1 | B1 not dependent on previous marks |
| **Alternative:** $P(\text{at most }7) = P(0,1,2,3,4,5,6,7)$ | M1 | Binomial term of form ${}^{10}C_x p^x(1-p)^{10-x}$, $0<p<1$ any $p$, $x\neq 10,0$ |
| $= (0.65)^{10} + {}^{10}C_1(0.35)^1(0.65)^9 + \ldots + {}^{10}C_7(0.35)^7(0.65)^3$ | A1 | Correct unsimplified answer or individual terms evaluated seen |
| $= 0.995$ | B1 | |
---
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1-(0.65)^n > 0.99$, $\ 0.01 > (0.65)^n$ | M1 | Equation or inequality with $(0.65)^n$ **and** $0.01$ or $(0.35)^n$ **and** $0.99$ only. (Note $1-0.99$ is equivalent to $0.01$ etc.) |
| $n > 10.69$ | M1 | Solving $a^n = c$, $0<a,c<1$ using logs **or** Trial and Error. If answer inappropriate, at least 2 trials required for Trial and Error M mark |
| smallest $n = 11$ | A1 | CAO |
---3 The probability that Janice will buy an item online in any week is 0.35 . Janice does not buy more than one item online in any week.\\
(i) Find the probability that, in a 10 -week period, Janice buys at most 7 items online.\\
(ii) The probability that Janice buys at least one item online in a period of $n$ weeks is greater than 0.99 . Find the smallest possible value of $n$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q3 [6]}}