题目

Problem

On any day, Kino travels to school by bus, by car or on foot with probabilities 0.2, 0.1 and 0.7 respectively. The probability that he is late when he travels by bus is $x$ . The probability that he is late when he travels by car is $2x$ and the probability that he is late when he travels on foot is 0.25.

The probability that, on a randomly chosen day, Kino is late is 0.235.

(a) Find the value of $x$ .

(3)

(b) Find the probability that, on a randomly chosen day, Kino travels to school by car given that he is not late.

(2)

题目中文翻译

任意一天，Kino 乘公共汽车、乘汽车或步行上学的概率分别为 0.2、0.1 和 0.7。他乘公共汽车时迟到的概率为 $x$ ，乘汽车时迟到的概率为 $2x$ ，步行时迟到的概率为 0.25。

随机选择一天，Kino 迟到的概率为 0.235。

(a) 求 $x$ 的值。

(3)

(b) 随机选择一天，已知 Kino 没有迟到，求他乘汽车上学的概率。

(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

迟到概率按三种交通方式加权求和，列一次方程。

答题过程

展开

By conditioning on the mode of transport,

\begin{align*} 0.2x+0.1(2x)+0.7(0.25)=&\,0.235,\\[4mm] 0.4x+0.175=&\,0.235,\\[4mm] x=&\,\boxed{0.15}. \end{align*}

(b)

解法一

思路

展开

分子是乘车且不迟到，分母是总不迟到概率。

答题过程

展开

Using (a), the probability of being late by car is $2x=0.30$ . Hence

\begin{align*} \mathrm{P}(\text{car}\mid\text{not late}) =&\,\frac{0.1(1-0.30)}{1-0.235}\\[4mm] =&\,\frac{0.07}{0.765}=\boxed{\frac{14}{153}}. \end{align*}

CIE 9709 2022 November Paper 52 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程