CAIE S1 2004 June — Question 3 5 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2004
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeConstruct probability distribution from scenario
DifficultyModerate -0.8 This is a straightforward probability distribution question requiring systematic enumeration of outcomes from two dice. Part (i) involves counting favorable outcomes (e.g., X=1 occurs in 11 cases out of 36), which is routine but slightly tedious. Part (ii) is a direct application of the expectation formula. No novel insight or complex problem-solving is required—just careful counting and basic probability calculations, making it easier than average.
Spec2.04a Discrete probability distributions

3 Two fair dice are thrown. Let the random variable \(X\) be the smaller of the two scores if the scores are different, or the score on one of the dice if the scores are the same.
  1. Copy and complete the following table to show the probability distribution of \(X\).
    \(x\)123456
    \(\mathrm { P } ( X = x )\)
  2. Find \(\mathrm { E } ( X )\).

Question 3:
Part (i)
AnswerMarks Guidance
\(x\)1 2
\(P(X=x)\)\(\frac{11}{36}\) \(\frac{9}{36}\)
M1For 36 in the uncancelled denominator somewhere; accept decimals e.g. 0.305 recurring or 0.306 etc
A1For 3 correct probabilities
A1All correct
Total: 3
Part (ii)
AnswerMarks Guidance
\(E(X) = 1 \times \frac{11}{36} + 2 \times \frac{9}{36} + 3 \times \frac{7}{36} + 4 \times \frac{5}{36} + 5 \times \frac{3}{36} + 6 \times \frac{1}{36} = \frac{91}{36}\)M1 For calculation of \(\sum xp\) where all probs \(< 1\)
A1
Total: 2
# Question 3:

## Part (i)

| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X=x)$ | $\frac{11}{36}$ | $\frac{9}{36}$ | $\frac{7}{36}$ | $\frac{5}{36}$ | $\frac{3}{36}$ | $\frac{1}{36}$ |

| M1 | For 36 in the uncancelled denominator somewhere; accept decimals e.g. 0.305 recurring or 0.306 etc
| A1 | For 3 correct probabilities
| A1 | All correct
**Total: 3**

## Part (ii)
$E(X) = 1 \times \frac{11}{36} + 2 \times \frac{9}{36} + 3 \times \frac{7}{36} + 4 \times \frac{5}{36} + 5 \times \frac{3}{36} + 6 \times \frac{1}{36} = \frac{91}{36}$ | M1 | For calculation of $\sum xp$ where all probs $< 1$
| A1 |
**Total: 2**

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3 Two fair dice are thrown. Let the random variable $X$ be the smaller of the two scores if the scores are different, or the score on one of the dice if the scores are the same.\\
(i) Copy and complete the following table to show the probability distribution of $X$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = x )$ &  &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

(ii) Find $\mathrm { E } ( X )$.

\hfill \mbox{\textit{CAIE S1 2004 Q3 [5]}}