CAIE S1 2020 June — Question 1 5 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeCompound event with two dice/coins
DifficultyModerate -0.8 This is a straightforward geometric distribution question with standard calculations. Part (a) is basic enumeration of outcomes, parts (b)-(d) apply standard geometric distribution formulas (mean = 1/p, P(X=k) = (1-p)^(k-1)p, cumulative probability). All steps are routine with no problem-solving insight required beyond recognizing the geometric distribution setup.
Spec2.03a Mutually exclusive and independent events5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)

1 The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.
  1. Show that the probability that the score is 4 is \(\frac { 1 } { 12 }\).
    The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is denoted by the random variable \(X\).
  2. Find the mean of \(X\).
  3. Find the probability that a score of 4 is first obtained on the 6th throw.
  4. Find \(\mathrm { P } ( X < 8 )\).

Question 1:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
Prob of 4 (from 1,3, 3,1 or 2,2) \(= \frac{3}{36} = \frac{1}{12}\)B1 AG
Total: 1
Part (b)
AnswerMarks Guidance
AnswerMark Guidance
Mean \(= \frac{1}{\frac{1}{12}} = 12\)B1
Total: 1
Part (c)
AnswerMarks Guidance
AnswerMark Guidance
\(\left(\frac{11}{12}\right)^5 \times \frac{1}{12} = 0.0539\) or \(\frac{161051}{2985984}\)B1
Total: 1
Part (d)
AnswerMarks Guidance
AnswerMark Guidance
\(1 - \left(\frac{11}{12}\right)^7\)M1
\(0.456\) or \(\frac{16344637}{35831808}\)A1
Total: 2
## Question 1:

**Part (a)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Prob of 4 (from 1,3, 3,1 or 2,2) $= \frac{3}{36} = \frac{1}{12}$ | B1 | AG |
| **Total: 1** | | |

---

**Part (b)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \frac{1}{\frac{1}{12}} = 12$ | B1 | |
| **Total: 1** | | |

---

**Part (c)**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{11}{12}\right)^5 \times \frac{1}{12} = 0.0539$ or $\frac{161051}{2985984}$ | B1 | |
| **Total: 1** | | |

---

**Part (d)**

| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - \left(\frac{11}{12}\right)^7$ | M1 | |
| $0.456$ or $\frac{16344637}{35831808}$ | A1 | |
| **Total: 2** | | |
1 The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that the score is 4 is $\frac { 1 } { 12 }$.\\

The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is denoted by the random variable $X$.
\item Find the mean of $X$.
\item Find the probability that a score of 4 is first obtained on the 6th throw.
\item Find $\mathrm { P } ( X < 8 )$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2020 Q1 [5]}}