CAIE S1 2010 June — Question 5 8 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Definitions
TypeDice probability problems
DifficultyModerate -0.8 This is a straightforward probability question requiring enumeration of outcomes for two dice, basic probability calculations (counting favorable outcomes over total outcomes), and applying definitions of exclusive and independent events. All parts involve routine application of standard probability concepts with no problem-solving insight needed—just careful counting and checking definitions.
Spec2.03a Mutually exclusive and independent events

Two fair twelve-sided dice with sides marked 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 are thrown, and the numbers on the sides which land face down are noted. Events \(Q\) and \(R\) are defined as follows. \(Q\): the product of the two numbers is 24. \(R\): both of the numbers are greater than 8.
  1. Find \(\mathrm{P}(Q)\). [2]
  2. Find \(\mathrm{P}(R)\). [2]
  3. Are events \(Q\) and \(R\) exclusive? Justify your answer. [2]
  4. Are events \(Q\) and \(R\) independent? Justify your answer. [2]

AnswerMarks Guidance
(i) \((2,12), (3, 8), (4, 6), (6, 4), (8,3), (12,2)\)M1 Listing or picking out at least 3 different options from a 12 by 12 (mult) table or seeing 3, 4, 5 or 6/144
\(P(Q) = \frac{6}{144} \text{ (1/24) (0.0417)}\)A1 [2] Correct answer
(ii) \(P(\text{both} > 8) = 1/3 \times 1/3 = 1/9 = P(R) \text{ (16/144)}\)M1, A1 [2] Squaring a sensible prob or picking out 12 – 25 options. Correct answer
(iii) since \(P(R \text{ and } Q) = 0\)
AnswerMarks Guidance
Yes, \(R\) and \(Q\) are exclusiveB1*, B1dep [2] o.e. in words
(iv) \(P(R \text{ and } Q) = 0 \neq P(R) \times P(Q)\)
AnswerMarks Guidance
or \(P(RQ) = 0 \neq P(R)\)
No, not independentB1*, B1dep [2] o.e. in words
(i) $(2,12), (3, 8), (4, 6), (6, 4), (8,3), (12,2)$ | M1 | Listing or picking out at least 3 different options from a 12 by 12 (mult) table or seeing 3, 4, 5 or 6/144

$P(Q) = \frac{6}{144} \text{ (1/24) (0.0417)}$ | A1 [2] | Correct answer

(ii) $P(\text{both} > 8) = 1/3 \times 1/3 = 1/9 = P(R) \text{ (16/144)}$ | M1, A1 [2] | Squaring a sensible prob or picking out 12 – 25 options. Correct answer

(iii) since $P(R \text{ and } Q) = 0$
Yes, $R$ and $Q$ are exclusive | B1*, B1dep [2] | o.e. in words

(iv) $P(R \text{ and } Q) = 0 \neq P(R) \times P(Q)$
or $P(R|Q) = 0 \neq P(R)$
No, not independent | B1*, B1dep [2] | o.e. in words

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Two fair twelve-sided dice with sides marked 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 are thrown, and the numbers on the sides which land face down are noted. Events $Q$ and $R$ are defined as follows.

$Q$: the product of the two numbers is 24.

$R$: both of the numbers are greater than 8.

\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm{P}(Q)$. [2]

\item Find $\mathrm{P}(R)$. [2]

\item Are events $Q$ and $R$ exclusive? Justify your answer. [2]

\item Are events $Q$ and $R$ independent? Justify your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2010 Q5 [8]}}