题目

Problem

Jasmine throws two ordinary fair 6-sided dice at the same time and notes the numbers on the uppermost faces. The events $A$ and $B$ are defined as follows.

$A$ : The sum of the two numbers is less than 6.

$B$ : The difference between the two numbers is at most 2.

(a) Determine whether or not the events $A$ and $B$ are independent.

(4)

(b) Find $\mathrm{P}(B\mid A')$ .

(3)

题目中文翻译

Jasmine 同时掷两枚普通的均匀六面骰子，并记录朝上两面的数字。事件 $A$ 和 $B$ 定义如下。

$A$ ：两个数字之和小于 6。

$B$ ：两个数字之差至多为 2。

(a) 判断事件 $A$ 和 $B$ 是否相互独立。

(4)

(b) 求 $\mathrm{P}(B\mid A')$ 。

(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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36个骰子有序结果等概率。分别数和小于6、差至多2及交集，再检验概率乘积。

答题过程

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There are 36 equally likely ordered outcomes.

For $A$ , sums 2, 3, 4 and 5 have $1+2+3+4=10$ outcomes. For $B$ , differences 0, 1 and 2 have $6+10+8=24$ outcomes. Thus

\mathrm{P}(A)=\frac{10}{36},\qquad\mathrm{P}(B)=\frac{24}{36}.

Among the ten outcomes in A, only $(1,4)$ and $(4,1)$ fail B. Hence

\mathrm{P}(A\cap B)=\frac8{36}.

But

\mathrm{P}(A)\mathrm{P}(B)=\frac{10}{36}\frac{24}{36} =\frac5{27}\neq\frac8{36}.

Therefore A and B are not independent.

(b)

解法一

思路

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B且非A的概率为B减去交集；条件分母是非A。

答题过程

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Using (a),

\begin{align*} \mathrm{P}(B\cap A')=&\,\frac{24-8}{36}=\frac{16}{36},\\[4mm] \mathrm{P}(A')=&\,1-\frac{10}{36}=\frac{26}{36}. \end{align*}

Hence

\mathrm{P}(B\mid A')=\frac{16/36}{26/36}=\boxed{\frac8{13}}.

CIE 9709 2023 June Paper 53 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