| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Geometric then binomial separate scenarios |
| Difficulty | Standard +0.3 This question tests standard applications of geometric and binomial distributions with straightforward calculations. Part (a) requires basic geometric distribution formulas (CDF and expectation), part (b) is a routine binomial probability calculation, and part (c) involves a standard normal approximation to binomial. All techniques are textbook exercises requiring recall and computation rather than problem-solving or insight, making it slightly easier than average. |
| Spec | 2.04c Calculate binomial probabilities2.04d Normal approximation to binomial5.02f Geometric distribution: conditions5.02g 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 |
|---|---|---|
| \([P(2 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) =] 1-(0.7)^6-(1-0.7)\) | M1 | \(1-0.7^n\) seen, \(n=5,6\) |
| \(= 0.582\) | A1 | www 0.582351 to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=2,3,4,5,6) = 0.7\times0.3+0.7^2\times0.3+0.7^3\times0.3+0.7^4\times0.3+0.7^5\times0.3\) \(= 0.21+0.147+0.1029+0.07203+0.050421\) | M1 | Sum of first 4 or 5 correct terms – no incorrect terms |
| \(= 0.582\) | A1 | www 0.582351 to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\frac{1}{3}\) | B1 | Condone 3.33, \(3.\dot{3}\) or \(\frac{10}{3}\) – NOT \(\frac{1}{0.3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(3,4,5) =]\ ^5C_3(0.3)^3(0.7)^2 + ^5C_4(0.3)^4(0.7)^1 + ^5C_5(0.3)^5(0.7)^0\) | M1 | One term seen \(^5C_x\,(p)^x(1-p)^{5-x}\), \(0 |
| \(= 0.1323+0.02835+0.00243\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(= 0.163,\ \frac{4077}{25000}\) | B1 | 0.16308 to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| \([1-P(0,1,2) =] 1-(^5C_0(0.3)^0(0.7)^5+^5C_1(0.3)^1(0.7)^4+^5C_2(0.3)^2(0.7)^3)\) | M1 | One term \(^5C_x\,(p)^x(1-p)^{5-x}\), \(0 |
| \(= 1-(0.16807+0.36015+0.3087)\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(= 0.163,\ \frac{4077}{25000}\) | B1 | 0.16308 to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Mean} = 75\times0.3 =]\ 22.5\) ; \([\text{Var} = 75\times0.3\times0.7 =]\ 15.75\) | B1 | 22.5, \(22\frac{1}{2}\) and 15.75, \(15\frac{3}{4}\) seen; \(\left(\sigma=\frac{3\sqrt{7}}{2}\right)\) or 3.9686269... to at least 3SF implies correct variance |
| \([P(X>20) =]\ P\!\left(Z > \frac{20.5-22.5}{\sqrt{15.75}}\right)\) | M1 | Substituting their \(\mu\) and \(\sigma\) into \(\pm\)standardisation formula, not \(\sigma^2\) not \(\sqrt{\sigma}\) |
| M1 | Using continuity correction 19.5 or 20.5 in *their* standardisation formula | |
| \([P(Z>-0.504) = \Phi(0.504)]\) \(= 0.693\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability. Expect final answer \(> 0.5\). Correct final answer implies this M1. |
| A1 | \(0.6925 < p \leq 0.693\) |
## Question 5(a)(i):
**Method 1:**
$[P(2 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) =] 1-(0.7)^6-(1-0.7)$ | M1 | $1-0.7^n$ seen, $n=5,6$
$= 0.582$ | A1 | www 0.582351 to at least 3SF
**Method 2:**
$P(X=2,3,4,5,6) = 0.7\times0.3+0.7^2\times0.3+0.7^3\times0.3+0.7^4\times0.3+0.7^5\times0.3$ $= 0.21+0.147+0.1029+0.07203+0.050421$ | M1 | Sum of first 4 or 5 correct terms – no incorrect terms
$= 0.582$ | A1 | www 0.582351 to at least 3SF
---
## Question 5(a)(ii):
$3\frac{1}{3}$ | B1 | Condone 3.33, $3.\dot{3}$ or $\frac{10}{3}$ – NOT $\frac{1}{0.3}$
---
## Question 5(b):
**Method 1:**
$[P(3,4,5) =]\ ^5C_3(0.3)^3(0.7)^2 + ^5C_4(0.3)^4(0.7)^1 + ^5C_5(0.3)^5(0.7)^0$ | M1 | One term seen $^5C_x\,(p)^x(1-p)^{5-x}$, $0<p<1$, $x\neq0,5$
$= 0.1323+0.02835+0.00243$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer
$= 0.163,\ \frac{4077}{25000}$ | B1 | 0.16308 to at least 3SF
**Method 2:**
$[1-P(0,1,2) =] 1-(^5C_0(0.3)^0(0.7)^5+^5C_1(0.3)^1(0.7)^4+^5C_2(0.3)^2(0.7)^3)$ | M1 | One term $^5C_x\,(p)^x(1-p)^{5-x}$, $0<p<1$, $x\neq0,5$
$= 1-(0.16807+0.36015+0.3087)$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer
$= 0.163,\ \frac{4077}{25000}$ | B1 | 0.16308 to at least 3SF
---
## Question 5(c):
$[\text{Mean} = 75\times0.3 =]\ 22.5$ ; $[\text{Var} = 75\times0.3\times0.7 =]\ 15.75$ | B1 | 22.5, $22\frac{1}{2}$ and 15.75, $15\frac{3}{4}$ seen; $\left(\sigma=\frac{3\sqrt{7}}{2}\right)$ or 3.9686269... to at least 3SF implies correct variance
$[P(X>20) =]\ P\!\left(Z > \frac{20.5-22.5}{\sqrt{15.75}}\right)$ | M1 | Substituting their $\mu$ and $\sigma$ into $\pm$standardisation formula, not $\sigma^2$ not $\sqrt{\sigma}$
| M1 | Using continuity correction 19.5 or 20.5 in *their* standardisation formula
$[P(Z>-0.504) = \Phi(0.504)]$ $= 0.693$ | M1 | Appropriate area $\Phi$, from final process, must be a probability. Expect final answer $> 0.5$. Correct final answer implies this M1.
| A1 | $0.6925 < p \leq 0.693$
---5 The probability that a driver passes an advanced driving test is 0.3 on any given attempt.
\begin{enumerate}[label=(\alph*)]
\item Dipak keeps taking the test until he passes. The random variable $X$ denotes the number of attempts required for Dipak to pass the test.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( 2 \leqslant X \leqslant 6 )$.
\item Find $\mathrm { E } ( X )$.\\
Five friends will each take their advanced driving test tomorrow.
\end{enumerate}\item Find the probability that at least three of them will pass tomorrow.\\
75 people will take their advanced driving test next week.\\[0pt]
\item Use an approximation to find the probability that more than 20 of them will pass next week. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q5 [11]}}