题目

Problem

Anil is taking part in a tournament. In each game in this tournament, players are awarded 2 points for a win, 1 point for a draw and 0 points for a loss. For each of Anil’s games, the probabilities that he will win, draw or lose are $0.5$ , $0.3$ and $0.2$ respectively. The results of the games are all independent of each other.

The random variable $X$ is the total number of points that Anil scores in his first 3 games in the tournament.

(a) Show that $\mathrm{P}(X=2)=0.114$ .

(2)

(b) Complete the probability distribution table for $X$ .

(3)

$x$	0	1	2	3	4	5	6
$\mathrm{P}(X=x)$			0.114	0.207	0.285		0.125

(3)

题目中文翻译

Anil 正在参加一项锦标赛。在每场比赛中，获胜得 2 分，平局得 1 分，失败得 0 分。Anil 在每场比赛中获胜、平局和失败的概率分别为 $0.5$ 、 $0.3$ 和 $0.2$ 。各场比赛的结果相互独立。

随机变量 $X$ 为 Anil 在锦标赛前三场比赛中的总得分。

(a) 证明 $\mathrm{P}(X=2)=0.114$ 。

(2)

(b) 补全 $X$ 的概率分布表。

(3)

$x$	0	1	2	3	4	5	6
$\mathrm{P}(X=x)$			0.114	0.207	0.285		0.125

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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总分2只有一胜两负或两平一负，每类有3种次序。

答题过程

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Let W, D and L denote a win, draw and loss. A total of 2 points requires WLL or DDL, in any order. Hence

\begin{align*} \mathrm{P}(X=2)=&\,3(0.5)(0.2)^2+3(0.3)^2(0.2)\\[4mm] =&\,0.060+0.054=\boxed{0.114}. \end{align*}

(b)

解法一

思路

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三个缺值分别对应三负、一平两负、两胜一平。

答题过程

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The missing probabilities are

\mathrm{P}(X=0)=(0.2)^3=0.008,

\mathrm{P}(X=1)=3(0.3)(0.2)^2=0.036,

\mathrm{P}(X=5)=3(0.5)^2(0.3)=0.225.

Thus the completed table is

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$\mathrm{P}(X=x)$	$0.008$	$0.036$	$0.114$	$0.207$	$0.285$	$0.225$	$0.125$

(c)

解法一

思路

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用完整分布表计算一阶与二阶矩，再求方差。

答题过程

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\begin{align*} \mathrm{E}(X)=&\,0.036+2(0.114)+3(0.207)\\[4mm] &\,\hspace{2pt}+4(0.285)+5(0.225)+6(0.125)\\[4mm] =&\,3.9. \end{align*}

Also,

\begin{align*} \mathrm{E}(X^2)=&\,0.036+4(0.114)+9(0.207)\\[4mm] &\,\hspace{2pt}+16(0.285)+25(0.225)+36(0.125)\\[4mm] =&\,17.04. \end{align*}

Therefore

\operatorname{Var}(X)=17.04-(3.9)^2=\boxed{1.83}.

CIE 9709 2024 March Paper 52 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程