题目
Problem
The Venn diagram shows the probability of a randomly selected student from a school being in the sets L, B and C, where
L represents the event that the student has instrumental music lessons
B represents the event that the student plays in the school band
C represents the event that the student sings in the school choir
p, q, r and s are probabilities.
(a) Select a pair of mutually exclusive events from L, B and C.
(1)
Given that P(L)=0.4, P(B)=0.13, P(C)=0.3 and the events L and C are independent,
(b) find the value of p,
(2)
(c) find the value of q, the value of r and the value of s.
(3)
A student is selected at random from those who play in the school band or sing in the school choir.
(d) Find the exact probability that this student has instrumental music lessons.
(3)
题目中文翻译
文氏图表示从某学校随机选取一名学生时，该学生属于集合 L、B、C 各区域的概率。其中，L 表示学生学习器乐，B 表示学生参加校乐队，C 表示学生参加校合唱团；p、q、r、s 均为概率。
(a) 从 L、B、C 中选出一对互斥事件。
已知 P(L)=0.4、P(B)=0.13、P(C)=0.3，且事件 L 与 C 相互独立。
(b) 求 p。
(c) 求 q、r 和 s。
从参加校乐队或校合唱团的学生中随机选取一人。
(d) 求该学生学习器乐的精确概率。
解答
(a)
解法一
思路
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从图中看，band 和 choir 没有重叠区域。
答题过程
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B and C are mutually exclusive.
(b)
解法一
思路
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独立事件满足 P(L∩C)=P(L)P(C)，而图中 L∩C 就是 p。
答题过程
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Since L and C are independent,
p=P(L∩C)=P(L)P(C)=0.4(0.3)=0.12.
(c)
解法一
思路
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分别用 P(L)、P(C) 和总概率为 1 来求剩下区域。
答题过程
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Since P(L)=0.4,
q+0.13+p=0.4.
So
q=0.4−0.13−0.12=0.15.
Since P(C)=0.3,
r+p=0.3,
so
r=0.3−0.12=0.18.
Finally,
s=1−(0.15+0.13+0.12+0.18)=0.42.
(d)
解法一
思路
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条件是 B∪C，分母是 band 或 choir 的总概率。分子是同时在 L 且属于 B∪C，即 0.13+p。
答题过程
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P(L∣B∪C)=P(B∪C)P(L∩(B∪C)).
The numerator is
P(L∩(B∪C))=0.13+0.12=0.25.
Since B and C are mutually exclusive,
P(B∪C)=0.13+0.30=0.43.
Therefore
P(L∣B∪C)=0.430.25=4325.
