| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Multiple independent binomial calculations |
| Difficulty | Moderate -0.5 This question involves three standard binomial/geometric distribution calculations with straightforward application of formulas. Part (a) is a direct binomial probability P(X < 8), part (b) is geometric distribution P(X ≤ 4), and part (c) is negative binomial. All are routine textbook exercises requiring formula recall and calculator work, but no problem-solving insight. Slightly easier than average due to the mechanical nature. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X < 8) = 1 - P(8, 9, 10) =]\) \(1 - (^{10}C_8(0.7)^8(0.3)^2 + ^{10}C_9(0.7)^9(0.3) + (0.7)^{10})\) | M1 | One term \(^{10}C_x(p)^x(1-p)^{10-x}\) with \(0 < p < 1\), \(x \neq 0\) or 10 |
| Correct unsimplified expression | A1 | Correct unsimplified expression. Condone omission of last bracket only |
| \(= [1 - (0.2335 + 0.1211 + 0.0282)] = 0.617\) | B1 | \(0.617 \leqslant p < 0.6175\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(0,1,2,3,4,5,6,7) =]\) \(0.3^{10} + ^{10}C_9\ 0.7\ 0.3^9 + \ldots + ^{10}C_3\ 0.7^7\ 0.3^3\) | M1 | One term \(^{10}C_x(p)^x(1-p)^{10-x}\) with \(0 < p < 1\), \(x \neq 0\) or 10 |
| Correct unsimplified expression | A1 | Correct unsimplified expression |
| \([= 5.905\times10^{-6} + 1.378\times10^{-3} + \ldots + 0.2668] = 0.617\) | B1 | \(0.617 \leqslant p < 0.6175\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X < 5) =]\ 1 - 0.3^4\) | M1 | \(1 - b^d\); \(b = 0.3, 0.7\); \(d = 4, 5\). \(1 - c^e - (1-c)\times c^{e-1}\); \(c = 0.3, 0.7\); \(e = 5, 6\) |
| \(0.9919,\ \dfrac{9919}{10000}\) | A1 | Condone 0.992. If M0 scored, SC B1 for 0.9919 or \(\dfrac{9919}{10000}\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X < 5) =]\ (0.7) + (0.3)(0.7) + (0.3)^2(0.7) + (0.3)^3(0.7)\) | M1 | \((e) + (f)(e) + (f)^2(e) + (f)^3(e)\ [+ (f)^4(e)]\); \(e = 0.7, 0.3\); \(e + f = 1\) |
| \(0.9919,\ \dfrac{9919}{10000}\) | A1 | Condone 0.992. If M0 scored, SC B1 for 0.9919 or \(\dfrac{9919}{10000}\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((0.4)^4(0.6)^2 \times 0.6 \times\ ^6C_2\) | M1 | \((0.4)^4(0.6)^r\); \(r = 2, 3\). No inappropriate addition |
| M1 | \((0.4)^a(0.6)^b \times\ ^6C_2\); \(a + b = 6, 7\) | |
| \(= 0.0829,\ \dfrac{1296}{15625}\) | A1 | Accept 0.082944 correct to at least 3 sig. figs. If A0 scored, SC B1 for correct answer |
## Question 2(a):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X < 8) = 1 - P(8, 9, 10) =]$ $1 - (^{10}C_8(0.7)^8(0.3)^2 + ^{10}C_9(0.7)^9(0.3) + (0.7)^{10})$ | M1 | One term $^{10}C_x(p)^x(1-p)^{10-x}$ with $0 < p < 1$, $x \neq 0$ or 10 |
| Correct unsimplified expression | A1 | Correct unsimplified expression. Condone omission of last bracket only |
| $= [1 - (0.2335 + 0.1211 + 0.0282)] = 0.617$ | B1 | $0.617 \leqslant p < 0.6175$ |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(0,1,2,3,4,5,6,7) =]$ $0.3^{10} + ^{10}C_9\ 0.7\ 0.3^9 + \ldots + ^{10}C_3\ 0.7^7\ 0.3^3$ | M1 | One term $^{10}C_x(p)^x(1-p)^{10-x}$ with $0 < p < 1$, $x \neq 0$ or 10 |
| Correct unsimplified expression | A1 | Correct unsimplified expression |
| $[= 5.905\times10^{-6} + 1.378\times10^{-3} + \ldots + 0.2668] = 0.617$ | B1 | $0.617 \leqslant p < 0.6175$ |
**Total: 3 marks**
---
## Question 2(b):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X < 5) =]\ 1 - 0.3^4$ | M1 | $1 - b^d$; $b = 0.3, 0.7$; $d = 4, 5$. $1 - c^e - (1-c)\times c^{e-1}$; $c = 0.3, 0.7$; $e = 5, 6$ |
| $0.9919,\ \dfrac{9919}{10000}$ | A1 | Condone 0.992. If M0 scored, SC B1 for 0.9919 or $\dfrac{9919}{10000}$ only |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X < 5) =]\ (0.7) + (0.3)(0.7) + (0.3)^2(0.7) + (0.3)^3(0.7)$ | M1 | $(e) + (f)(e) + (f)^2(e) + (f)^3(e)\ [+ (f)^4(e)]$; $e = 0.7, 0.3$; $e + f = 1$ |
| $0.9919,\ \dfrac{9919}{10000}$ | A1 | Condone 0.992. If M0 scored, SC B1 for 0.9919 or $\dfrac{9919}{10000}$ only |
**Total: 2 marks**
---
## Question 2(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(0.4)^4(0.6)^2 \times 0.6 \times\ ^6C_2$ | M1 | $(0.4)^4(0.6)^r$; $r = 2, 3$. No inappropriate addition |
| | M1 | $(0.4)^a(0.6)^b \times\ ^6C_2$; $a + b = 6, 7$ |
| $= 0.0829,\ \dfrac{1296}{15625}$ | A1 | Accept 0.082944 correct to at least 3 sig. figs. If A0 scored, SC B1 for correct answer |
**Total: 3 marks**
---2 Sam is a member of a soccer club. She is practising scoring goals. The probability that Sam will score a goal on any attempt is 0.7 , independently of all other attempts.
\begin{enumerate}[label=(\alph*)]
\item Sam makes 10 attempts at scoring goals.
Find the probability that Sam will score goals on fewer than 8 of these attempts.
\item Find the probability that Sam's first successful attempt will be before her 5th attempt.
\item Wei is a member of the same soccer club. He is also practising scoring goals. The probability that Wei will score a goal on any attempt is 0.6 , independently of all other attempts.
Wei is going to keep making attempts until he scores 3 goals.\\
Find the probability that he scores his third goal on his 7th attempt.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q2 [8]}}