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CIE 9709 2023 June Paper 53 Q1

A Level / CIE / S1

CIE 9709 2023 June Paper 53 Paper · Question 1

题目

Problem

Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable XX.

(a) State the value of E(X)\mathrm{E}(X).

(1)

(b) Find the probability that exactly 5 throws are required to obtain a pair of heads.

(1)

(c) Find the probability that fewer than 7 throws are required to obtain a pair of heads.

(2)

题目中文翻译

反复同时抛掷两枚均匀硬币,直到两枚都出现正面为止。用随机变量 XX 表示所需的抛掷次数。

(a) 写出 E(X)\mathrm{E}(X) 的值。

(1)

(b) 求恰好需要抛掷 5 次才得到两个正面的概率。

(1)

(c) 求抛掷少于 7 次便得到两个正面的概率。

(2)

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

一次两正面的成功概率1/4,几何分布均值为其倒数。

答题过程

展开

Each throw gives a pair of heads with probability 1212=14\frac12\frac12=\frac14. Thus

E(X)=11/4=4.\mathrm{E}(X)=\frac1{1/4}=\boxed{4}.

(b)

解法一

思路

展开

前四次未获两正面,第五次获两正面。

答题过程

展开 P(X=5)=(34)414=811024≈0.0791.\mathrm{P}(X=5)=\biggl(\frac34\biggr)^4\frac14 =\frac{81}{1024}\approx\boxed{0.0791}.

(c)

解法一

思路

展开

少于7即前6次内成功,取六次全失败的补集。

答题过程

展开 P(X<7)=1−(34)6=33674096≈0.822.\mathrm{P}(X<7)=1-\biggl(\frac34\biggr)^6 =\frac{3367}{4096}\approx\boxed{0.822}.

解法二

思路

展开

也可按首次两正面发生于第1至第6次分类,求等比和。

答题过程

展开 P(X<7)= ∑r=16(34)r−114= 141−(3/4)61−3/4= 33674096≈0.822.\begin{align*} \mathrm{P}(X<7) =&\,\sum_{r=1}^{6}\biggl(\frac34\biggr)^{r-1}\frac14\\[4mm] =&\,\frac14\frac{1-(3/4)^6}{1-3/4}\\[4mm] =&\,\frac{3367}{4096}\approx\boxed{0.822}. \end{align*}