CAIE S1 2023 June — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeCompound event with two dice/coins
DifficultyModerate -0.3 This is a straightforward geometric distribution question with standard probability p=1/4. Part (a) requires recall of E(X)=1/p, part (b) is direct substitution into P(X=5)=(3/4)^4(1/4), and part (c) requires summing a short geometric series or using the CDF formula. All parts are routine applications of standard formulas with no conceptual challenges, making it slightly easier than average.
Spec5.02f 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

1 Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable \(X\).
  1. State the value of \(\mathrm { E } ( X )\).
  2. Find the probability that exactly 5 throws are required to obtain a pair of heads.
  3. Find the probability that fewer than 7 throws are required to obtain a pair of heads.

Question 1:
Part (a)
AnswerMarks Guidance
\(\left[P(HH) = \frac{1}{4}\right]\) \([E(X) = 4]\)B1
*Total: 1 mark*
Part (b)
AnswerMarks Guidance
\(\left[P(X=5) = \left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)\right] = 0.0791\)B1 \(\frac{81}{1024}\)
*Total: 1 mark*
Part (c)
\([P(X < 7) =] 1 - \left(\frac{3}{4}\right)^6\)
or \(\frac{1}{4} + \frac{3}{4} \times \frac{1}{4} + \frac{3^2}{4} \times \frac{1}{4} + \ldots + \frac{3^5}{4} \times \frac{1}{4}\)
AnswerMarks Guidance
M1\(1 - p^n,\ 0 < p < 1,\ n = 6, 7\); or \(p + p(1-p) + p(1-p)^2 + \ldots + p(1-p)^n\), where \(n = 4, 5\)
\(= \frac{3367}{4096},\ 0.822\)A1 Accept \(0.82202148\ldots\) to at least 3SF
*Total: 2 marks*
## Question 1:

**Part (a)**

$\left[P(HH) = \frac{1}{4}\right]$ $[E(X) = 4]$ | **B1** | —

*Total: 1 mark*

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**Part (b)**

$\left[P(X=5) = \left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)\right] = 0.0791$ | **B1** | $\frac{81}{1024}$

*Total: 1 mark*

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**Part (c)**

$[P(X < 7) =] 1 - \left(\frac{3}{4}\right)^6$

or $\frac{1}{4} + \frac{3}{4} \times \frac{1}{4} + \frac{3^2}{4} \times \frac{1}{4} + \ldots + \frac{3^5}{4} \times \frac{1}{4}$

| **M1** | $1 - p^n,\ 0 < p < 1,\ n = 6, 7$; or $p + p(1-p) + p(1-p)^2 + \ldots + p(1-p)^n$, where $n = 4, 5$

$= \frac{3367}{4096},\ 0.822$ | **A1** | Accept $0.82202148\ldots$ to at least 3SF

*Total: 2 marks*
1 Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item State the value of $\mathrm { E } ( X )$.
\item Find the probability that exactly 5 throws are required to obtain a pair of heads.
\item Find the probability that fewer than 7 throws are required to obtain a pair of heads.
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2023 Q1 [4]}}