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工程数学概率论统计简明教程第二版复习大纲
2022-06-26 09:42:00 【JIeJaitt】
工程数学概率论统计简明教程第二版复习大纲
第二章 事件的概率
掷两颗骰子,求下列事件的概率:
(1) 点数之和为7;
(2) 点数之和不超过 5;
(3)点数之和为偶数。
分别记题(1)、(2)、(3)的事件为 A,B,C,样本点总数 n = 6 2 n=6^2 n=62
* A含样本点 ( 2 , 5 ) (2,5) (2,5), ( 5 , 2 ) (5,2) (5,2), ( 1 , 6 ) (1,6) (1,6), ( 6 , 1 ) (6,1) (6,1), ( 3 , 4 ) (3,4) (3,4), ( 4 , 3 ) (4,3) (4,3)
∴ P ( A ) = 6 6 2 = 1 6 \therefore P(A)=\frac {6}{6^2}=\frac {1}{6} ∴P(A)=626=61
* B含样本点 ( 1 , 1 ) (1,1) (1,1), ( 1 , 2 ) (1,2) (1,2), ( 2 , 1 ) (2,1) (2,1), ( 1 , 3 ) , (1,3), (1,3), ( 3 , 1 ) (3,1) (3,1), ( 1 , 4 ) (1,4) (1,4), ( 4 , 1 ) (4,1) (4,1), ( 2 , 2 ) (2,2) (2,2), ( 2 , 3 ) (2,3) (2,3), ( 3 , 2 ) (3,2) (3,2)
∴ P ( B ) = 10 6 2 = 5 18 \therefore P(B)=\frac {10}{6^2}=\frac {5}{18} ∴P(B)=6210=185
* C含样本点 ( 1 , 1 ) (1,1) (1,1), ( 1 , 3 ) (1,3) (1,3), ( 3 , 1 ) (3,1) (3,1), ( 1 , 5 ) (1,5) (1,5), ( 5 , 1 ) (5,1) (5,1), ( 2 , 2 ) (2,2) (2,2), ( 2 , 4 ) (2,4) (2,4), ( 4 , 2 ) (4,2) (4,2), ( 2 , 6 ) (2,6) (2,6), ( 6 , 2 ) (6,2) (6,2), ( 3 , 3 ) (3,3) (3,3), ( 3 , 5 ) (3,5) (3,5), ( 5 , 3 ) (5,3) (5,3), ( 4 , 4 ) (4,4) (4,4), ( 4 , 6 ) (4,6) (4,6), ( 6 , 4 ) (6,4) (6,4), ( 5 , 5 ) (5,5) (5,5), ( 6 , 6 ) (6,6) (6,6),一共18个样本点。
∴ P ( B ) = 18 6 2 = 1 2 \therefore P(B)=\frac {18}{6^2}=\frac {1}{2} ∴P(B)=6218=21甲、乙两艘轮船都要在某个泊位停靠 6 h 6h 6h,假定它们在一昼夜的时间段中随机地到达,试求这两艘船中至少有一艘在停靠泊位时必须等待的概率.


已知 A A A ⊂ \subset ⊂ B B B, P ( A ) P(A) P(A) ⊂ \subset ⊂= 0.4 0.4 0.4, P ( B ) P(B) P(B)= 0.6 0.6 0.6,求
(1) P ( A ‾ ) P(\overline{A}) P(A), P ( B ‾ ) P(\overline{B}) P(B);
(2) P ( A ∪ B ) P(A\cup B) P(A∪B);
(3) P ( A B ) P(AB) P(AB);
(4) P ( B ‾ A ) P(\overline{B}A) P(BA), P ( A B ‾ ) P(\overline{AB}) P(AB);
(5) P ( A ‾ B ) P(\overline{A}B) P(AB)。
* P ( A ‾ ) = 1 − P ( A ) = 1 − 0.4 = 0.6 P ( B ‾ ) = 1 − P ( B ) = 1 − 0.6 = 0.4 \begin{aligned} &P(\overline{A}) =1- P(A) =1-0.4 = 0.6 \\ &P(\overline{B}) =1- P(B) =1-0.6=0.4 \end{aligned} P(A)=1−P(A)=1−0.4=0.6P(B)=1−P(B)=1−0.6=0.4
* P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A B ) = P ( A ) + P ( B ) − P ( A ) = P ( B ) = 0.6 \begin{aligned} P(A\cup B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A) = P(B) = 0.6 \end{aligned} P(A∪B)=P(A)+P(B)−P(AB)=P(A)+P(B)−P(A)=P(B)=0.6
* P ( A B ) = P ( A ) = 0.4 \begin{aligned} P(AB) = P(A) = 0.4 \end{aligned} P(AB)=P(A)=0.4
* P ( B ‾ A ) = P ( A − B ) = P ( ϕ ) = 0 P ( A B ‾ ) = P ( A ∪ B ‾ ) = 1 − P ( A ∪ B ) = 1 − 0.6 = 0.4 \begin{aligned} &P(\overline{B}A)= P(A - B)= P(\phi)=0 \\ &P(\overline{AB})=P(\overline{A\cup B})=1- P(A\cup B)=1-0.6 = 0.4 \end{aligned} P(BA)=P(A−B)=P(ϕ)=0P(AB)=P(A∪B)=1−P(A∪B)=1−0.6=0.4
* P ( A ‾ B ) = P ( B − A ) = 0.6 − 0.4 = 0.2 \begin{aligned} P(\overline{A}B)= P(B - A)= 0.6 - 0.4 = 0.2 \end{aligned} P(AB)=P(B−A)=0.6−0.4=0.2设 A , B A,B A,B是两个事件,己知P(A)=0.5

第三章 条件概率和事件的独立性
- 从次品率为 p = 0.2 p=0.2 p=0.2的一批产品中,有放回抽取 5 次,每次取一件,分别求抽到的5件中恰好有了件次品以及至多有3件次品这两个事件的概率。

- 某工厂有甲、乙、丙三个车间,生产同一产品,每个车间的产量分别占全厂的 25%,35%,40%,各车间产品的次品率分别为 5%,4%,2%,求该厂产品的次品率。

- 总经理的五位秘书中有两位精通英语,今偶遇其中的三位秘书,求下列
事件的概率:
(1)事件A={其中恰有一位精通英语};
(2)事件B={其中恰有两位精通英语|;
(3)事件C={其中有人精通英语}.






第四章 随机变量及其分布







第五章 二维随机变量及其分布
第六章 随机变量的函数及其分布
第七章 随机变量的数字特征
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