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CIE 9709 2023 November Paper 51 Q5

A Level / CIE / S1

CIE 9709 2023 Nov Paper 51 Paper · Question 5

题目

Problem

A red spinner has four sides labelled 1, 2, 3, 4. When the spinner is spun, the score is the number on the side on which it lands. The random variable XX denotes this score. The probability distribution table for XX is given below.

xx1234
P(X=x)\mathrm{P}(X=x)0.28pp2p2p3p3p

(a) Show that p=0.12p=0.12.

(1)

A fair blue spinner and a fair green spinner each have four sides labelled 1, 2, 3, 4. All three spinners (red, blue and green) are spun at the same time.

(b) Find the probability that the sum of the three scores is 4 or less.

(3)

(c) Find the probability that the product of the three scores is 4 or less given that XX is odd.

(5)
题目中文翻译

一个红色转子有四个侧面，分别标有 1、2、3、4。转动转子后，得分为其着地侧面上的数字。随机变量 XX 表示这个得分。XX 的概率分布表如下。

xx1234
P(X=x)\mathrm{P}(X=x)0.28pp2p2p3p3p

(a) 证明 p=0.12p=0.12。

(1)

一个均匀的蓝色转子和一个均匀的绿色转子各有四个侧面，分别标有 1、2、3、4。同时转动这三个转子（红色、蓝色和绿色）。

(b) 求三个得分之和不超过 4 的概率。

(3)

(c) 已知 XX 为奇数，求三个得分之积不超过 4 的概率。

(5)
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

概率总和必须为1，先建立关于p的方程再求解。

答题过程

展开 0.28+p+2p+3p= 1,6p= 0.72,p= 0.12.\begin{align*} 0.28+p+2p+3p=&\,1,\\[4mm] 6p=&\,0.72,\\[4mm] p=&\,\boxed{0.12}. \end{align*}

(b)

解法一

思路

展开

最小总和是3，要总和不超过4只能是111或三个位置之一为2，红转子概率要单独对应。

答题过程

展开

In the order (red, blue, green), the permitted outcomes are (1,1,1)(1,1,1), (1,1,2)(1,1,2), (1,2,1)(1,2,1) and (2,1,1)(2,1,1).

The blue and green scores each have probability 14\frac14. Hence

P= 3(0.28)(14)2+0.12(14)2= 0.0525+0.0075=0.06.\begin{align*} \mathrm{P}=&\,3(0.28)\biggl(\frac14\biggr)^2 +0.12\biggl(\frac14\biggr)^2\\[4mm] =&\,0.0525+0.0075=\boxed{0.06}. \end{align*}

(c)

解法一

思路

展开

已知红转子为奇数，只可能1或3。红为1时蓝绿乘积至多4有八对；红为3时只能蓝绿均1。

答题过程

展开

Let AA denote a product of at most 4, and OO the event that the red score is odd. From (a),

P(O)=0.28+2(0.12)=0.52.\mathrm{P}(O)=0.28+2(0.12)=0.52.

If the red score is 1, the permitted ordered blue–green pairs are

(1,1), (1,2), (1,3), (1,4),(1,1),\ (1,2),\ (1,3),\ (1,4), (2,1), (2,2), (3,1), (4,1).(2,1),\ (2,2),\ (3,1),\ (4,1).

If the red score is 3, the only permitted pair is (1,1)(1,1). Therefore

P(A∩O)=0.28816+0.24116=0.155,\mathrm{P}(A\cap O)=0.28\frac8{16}+0.24\frac1{16}=0.155,

so

P(A∣O)=0.1550.52=31104.\mathrm{P}(A\mid O)=\frac{0.155}{0.52}=\boxed{\frac{31}{104}}.