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CIE 9709 2024 June Paper 52 Q5

A Level / CIE / S1

CIE 9709 2024 June Paper 52 Paper · Question 5

题目

Problem

Jasmine has one $5 coin, two $2 coins and two $1 coins. She selects two of these coins at random. The random variable XX is the total value, in dollars, of these two coins.

(a) Show that P(X=7)=0.2\mathrm{P}(X=7)=0.2.

(1)

(b) Draw up the probability distribution table for XX.

(3)

(c) Find the value of Var⁡(X)\operatorname{Var}(X).

(3)
题目中文翻译

Jasmine 有一枚 $5 硬币、两枚 $2 硬币和两枚 $1 硬币。她从这些硬币中随机选取两枚。随机变量 XX 为这两枚硬币的总面值,以元为单位。

(a) 证明 P(X=7)=0.2\mathrm{P}(X=7)=0.2。

(1)

(b) 列出 XX 的概率分布表。

(3)

(c) 求 Var⁡(X)\operatorname{Var}(X) 的值。

(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

展开

总额7只能是一枚5元与一枚2元,共两种实物硬币组合,总组合10种。

答题过程

展开

A total of 7 dollars requires the 5-dollar coin and one of the two 2-dollar coins. Therefore

P(X=7)=(11)(21)(52)=210=0.2.\mathrm{P}(X=7)=\frac{\binom11\binom21}{\binom52} =\frac2{10}=\boxed{0.2}.

解法二

思路

展开

另一条路线按抽取次序,先5后2或先2后5两条路径相加。

答题过程

展开 P(X=7)=1524+2514=220+220=0.2.\mathrm{P}(X=7)=\frac15\frac24+\frac25\frac14 =\frac2{20}+\frac2{20}=\boxed{0.2}.

(b)

解法一

思路

展开

五枚实际硬币选两枚共有10种等概率组合,逐个总额计数。

答题过程

展开

The possible coin values and numbers of selections are

Total xxCoin valuesSelections
221+11+111
331+21+22×2=42\times2=4
442+22+211
661+51+522
772+52+522

Dividing by (52)=10\binom52=10 gives

xx2233446677
P(X=x)\mathrm{P}(X=x)0.10.10.40.40.10.10.20.20.20.2

(c)

解法一

思路

展开

先求均值及二阶矩,方差为二阶矩减去均值平方。

答题过程

展开 E(X)= 2(0.1)+3(0.4)+4(0.1) +6(0.2)+7(0.2)= 4.4,\begin{align*} \mathrm{E}(X)=&\,2(0.1)+3(0.4)+4(0.1)\\[4mm] &\,\hspace{2pt}+6(0.2)+7(0.2)\\[4mm] =&\,4.4, \end{align*}

and

E(X2)= 22(0.1)+32(0.4)+42(0.1) +62(0.2)+72(0.2)=22.6.\begin{align*} \mathrm{E}(X^2)=&\,2^2(0.1)+3^2(0.4)+4^2(0.1)\\[4mm] &\,\hspace{2pt}+6^2(0.2)+7^2(0.2)=22.6. \end{align*}

Hence

Var⁡(X)=22.6−(4.4)2=3.24.\operatorname{Var}(X)=22.6-(4.4)^2=\boxed{3.24}.