当前位置:网站首页>Concise course of probability theory and statistics in engineering mathematics second edition review outline

Concise course of probability theory and statistics in engineering mathematics second edition review outline

2022-06-26 10:37:00 JIeJaitt

Chapter two The probability of the event

  1. Roll two dice , Find the probability of the following events :
    (1) The sum of the points is 7;
    (2) The sum of points does not exceed 5;
    (3) The sum of points is even .
    Write down the questions separately (1)、(2)、(3) Events for A,B,C, Total number of sample points n = 6 2 n=6^2 n=62
    * A Including sample points ( 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 Including sample points ( 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 Including sample points ( 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), altogether 18 A sample points .
    ∴ P ( B ) = 18 6 2 = 1 2 \therefore P(B)=\frac {18}{6^2}=\frac {1}{2} P(B)=6218=21

  2. nail 、 B both ships have to call at a certain berth 6 h 6h 6h, Suppose they arrive randomly in a day and night period , Try to find out the probability that at least one of the two ships must wait when berthing .
     Insert picture description here
     Insert picture description here

  3. It is known that 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, seek
    (1) P ( A ‾ ) P(\overline{A}) P(A), P ( B ‾ ) P(\overline{B}) P(B);
    (2) P ( A ∪ B ) P(A\cup B) P(AB);
    (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)=1P(A)=10.4=0.6P(B)=1P(B)=10.6=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(AB)=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(AB)=P(ϕ)=0P(AB)=P(AB)=1P(AB)=10.6=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(BA)=0.60.4=0.2

  4. set up A , B A,B A,B It's two events , Already know P(A)=0.5 Please add a picture description

The third chapter Conditional probability and independence of events

  1. The defective rate is p = 0.2 p=0.2 p=0.2 In a batch of products , Yes, put it back 5 Time , One at a time , Respectively find the 5 There are just pieces of defective products and at most 3 The probability of two events of defective products .
     Insert picture description here
  2. A factory has a 、 B 、 C three workshops , Produce the same product , The output of each workshop accounts for 25%,35%,40%, The defective rate of products in each workshop is 5%,4%,2%, Find the defective rate of the factory's products .
     Insert picture description here
  3. Two of the five secretaries of the general manager are proficient in English , I ran into three of the secretaries , Ask for the following
    The probability of the event :
    (1) event A={ One of them is proficient in English };
    (2) event B={ Two of them are proficient in English |;
    (3) event C={ Some of them are proficient in English }.

 Insert picture description here
 Insert picture description here
 Insert picture description here

 Insert picture description here
 Insert picture description here
 Insert picture description here

Chapter four Random variables and their distribution

 Insert picture description here
 Insert picture description here

 Insert picture description here

 Insert picture description here

 Insert picture description here
 Insert picture description here
 Insert picture description here

The fifth chapter Two dimensional random variables and their distribution

Chapter six Function and distribution of random variables

Chapter vii. The numerical characteristics of random variables

原网站

版权声明
本文为[JIeJaitt]所创,转载请带上原文链接,感谢
https://yzsam.com/2022/177/202206260941561748.html