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Concise course of probability theory and statistics in engineering mathematics second edition review outline
2022-06-26 10:37:00 【JIeJaitt】
Concise course of probability theory and statistics in engineering mathematics second edition review outline
- Chapter two The probability of the event
- The third chapter Conditional probability and independence of events
- Chapter four Random variables and their distribution
- 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
Chapter two The probability of the event
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=21nail 、 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 .


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(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=0.2set up A , B A,B A,B It's two events , Already know P(A)=0.5

The third chapter Conditional probability and independence of events
- 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 .

- 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 .

- 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 }.






Chapter four Random variables and their distribution







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
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