题目
Problem
A bag contains a large number of coloured counters. Each counter is labelled A, B or C
30% of the counters are labelled A
45% of the counters are labelled B
The rest of the counters are labelled C
It is known that
2% of the counters labelled A are red
4% of the counters labelled B are red
6% of the counters labelled C are red
One counter is selected at random from the bag.
(a) Complete the tree diagram on the opposite page to illustrate this information.
(2)
(b) Calculate the probability that the counter is labelled A and is not red.
(2)
(c) Calculate the probability that the counter is red.
(2)
(d) Given that the counter is red, find the probability that it is labelled C
(3)
题目中文翻译
一个袋子中装有大量彩色计数片，每片标有 A、B 或 C。30% 标有 A，45% 标有 B，其余标有 C。
已知标有 A 的计数片中有 2% 是红色，标有 B 的计数片中有 4% 是红色，标有 C 的计数片中有 6% 是红色。从袋中随机取出一片计数片。
(a) 补全另一页的树状图，以表示这些信息。
(b) 计算所取计数片标有 A 且不是红色的概率。
(c) 计算所取计数片是红色的概率。
(d) 已知所取计数片是红色，求它标有 C 的概率。
解答
(a)
解法一
思路
展开
第一层标签概率为 0.30,0.45,0.25。第二层每个标签下红色概率分别为 0.02,0.04,0.06，不红就是补到 1。
答题过程
展开
The missing label probabilities are
P(B)=0.45,P(C)=1−0.30−0.45=0.25.
The missing conditional probabilities are
P(not red∣A)=0.98,P(red∣B)=0.04,P(not red∣B)=0.96,P(red∣C)=0.06,P(not red∣C)=0.94.
(b)
解法一
思路
展开
沿树图的 A then not red 路线相乘。
答题过程
展开
P(A∩not red)=0.30(0.98)=0.294.
(c)
解法一
思路
展开
红色可以来自 A、B、C 三条路线，把三条路线的概率相加。
答题过程
展开
P(red)===0.30(0.02)+0.45(0.04)+0.25(0.06)0.006+0.018+0.0150.039.
(d)
解法一
思路
展开
这是条件概率：分子是“C 且 red”，分母是全部 red 的概率。
答题过程
展开
P(C∣red)====P(red)P(C∩red)0.0390.25(0.06)0.0390.015135.
So the probability is
0.385
to 3 significant figures.
