题目

Problem

A bag contains 4 blue marbles and 12 red marbles. One marble is selected at random from the bag. If this marble is blue, it is replaced in the bag, but if it is red, it is not replaced. A second marble is now selected at random from the bag.

(a) Find the probability that both marbles selected are the same colour.

(2)

(b) Find the probability that the first marble is blue given that the second marble is red.

(3)

题目中文翻译

一个袋子里有 4 颗蓝色弹珠和 12 颗红色弹珠。从袋中随机抽取一颗弹珠。如果这颗弹珠是蓝色的，就将它放回袋中；如果是红色的，就不放回。现在再从袋中随机抽取第二颗弹珠。

(a) 求抽到的两颗弹珠颜色相同的概率。

(2)

(b) 已知第二颗弹珠是红色的，求第一颗弹珠是蓝色的概率。

(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

先蓝时有放回，第二次仍有16颗；先红时不放回，第二次有15颗且红珠剩11颗。两条同色路径相加。

答题过程

展开

The same-colour outcomes are BB and RR. The first blue marble is replaced, whereas the first red is not. Thus

\begin{align*} \mathrm{P}(\text{same colour}) =&\,\frac4{16}\frac4{16}+\frac{12}{16}\frac{11}{15}\\[4mm] =&\,\boxed{\frac{49}{80}}. \end{align*}

(b)

解法一

思路

展开

分子只有蓝红路径，分母须包括蓝红和红红两条路径。

答题过程

展开

Write $B_1$ for first marble blue and $R_2$ for second marble red. Then

\mathrm{P}(B_1\cap R_2)=\frac4{16}\frac{12}{16}=\frac3{16},

and

\begin{align*} \mathrm{P}(R_2)=&\,\frac4{16}\frac{12}{16}+\frac{12}{16}\frac{11}{15}\\[4mm] =&\,\frac3{16}+\frac{11}{20}=\frac{59}{80}. \end{align*}

Hence

\mathrm{P}(B_1\mid R_2)=\frac{3/16}{59/80}=\boxed{\frac{15}{59}}.

CIE 9709 2025 June Paper 52 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程