题目

Problem

Freddie has two bags of marbles.

Bag $X$ contains 7 red marbles and 3 blue marbles.

Bag $Y$ contains 4 red marbles and 1 blue marble.

Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from.

(a) Draw a tree diagram to represent this information, showing the probability on each of the branches.

(3)

(b) Find the probability that both of the marbles removed from the bag are the same colour.

(4)

(2)

题目中文翻译

Freddie 有两袋弹珠。

袋 $X$ 中有 7 颗红色弹珠和 3 颗蓝色弹珠。

袋 $Y$ 中有 4 颗红色弹珠和 1 颗蓝色弹珠。

Freddie 随机选择其中一袋，从该袋中随机取出一颗弹珠且不放回。随后向每一袋中各加入一颗新的红色弹珠，再从取出第一颗弹珠的同一袋中随机取出第二颗弹珠。

(a) 画一幅树状图表示这些信息，并标出每条分支上的概率。

(3)

(b) 求从袋中取出的两颗弹珠颜色相同的概率。

(4)

(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

先选袋，再取第一枚。取出后补红，所选袋总数恢复原数；若第一枚蓝，则蓝数减1、红数加1。

答题过程

展开

Start with bag branches $X,Y$ , each of probability $\frac12$ . The next branches and final conditional probabilities are

Bag	First marble	Probability of first marble	Second R	Second B
$X$	R	$7/10$	$7/10$	$3/10$
$X$	B	$3/10$	$8/10$	$2/10$
$Y$	R	$4/5$	$4/5$	$1/5$
$Y$	B	$1/5$	$1$	$0$

In particular, removing the only blue marble from Y and adding a red marble leaves only red marbles in Y.

(b)

解法一

思路

展开

同色路径为XRR、XBB、YRR；YBB概率为零，不能加入非零项。

答题过程

展开

\begin{align*} \mathrm{P}(\text{same colour}) =&\,\frac12\frac7{10}\frac7{10}\\[4mm] &\,\hspace{2pt}+\frac12\frac3{10}\frac2{10}\\[4mm] &\,\hspace{4pt}+\frac12\frac45\frac45\\[4mm] =&\,\frac{49+6+64}{200}=\boxed{\frac{119}{200}}. \end{align*}

(c)

解法一

思路

展开

已知异色的分母可由(b)补集求；分子为YRB或YBR两条路径。

答题过程

展开

Using (b),

\mathrm{P}(\text{different colours})=1-\frac{119}{200}=\frac{81}{200}.

Also,

\mathrm{P}(Y\cap\text{different colours}) =\frac12\frac45\frac15+\frac12\frac15(1)=\frac9{50}.

Hence

\mathrm{P}(Y\mid\text{different colours}) =\frac{9/50}{81/200}=\boxed{\frac49}.

CIE 9709 2023 November Paper 52 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程