The student solves a difficult problem about the combinatorial analysis of probability theory

a33 (3 in the upper right and lower right of A) A33 (3 in the upper right and lower right of A) A22

a33×a44

A33 C133 (top right and bottom right of C) C13

As follows:

Due to the nature of the Poisson distribution (additive), the number of emergency calls received from 12 o'clock to 3 p.m. x obeys the Poisson distribution with the parameter (1 2) 3=, so the probability of not receiving a call for help from 12 noon to 3 p.m. on a given day p(k=0)=e (.

The number of emergency calls received from 12 p.m. to 5 p.m. x obeys the Poisson distribution with the parameter (1 2) 5=, so the probability of receiving at least one call for help from 12 noon to 5 p.m. on a given day p(k<=0)=p(k=0)+p(k=1)=e ().

Application examples. Poisson distributions are suitable for describing the number of occurrences of random events per unit of time (or space). For example, the number of people who arrive at a certain service facility within a certain period of time, the number of calls received by the switch, the number of waiting customers at the bus station, the number of machine failures, the number of natural disasters, the number of defects on a product, the number of bacteria distribution in the unit partition under the microscope, and so on.

Due to the nature of the Poisson distribution (additive), the number of emergency calls received from 12 o'clock to 3 p.m. x obeys the Poisson distribution with the parameter (1 2) 3=, so the probability of not receiving a call for help from 12 noon to 3 p.m. on a given day p(k=0)=e (.

The number of emergency calls received from 12 p.m. to 5 p.m. x obeys the Poisson distribution with the parameter (1 2) 5=, so the probability of receiving at least one call for help from 12 noon to 5 p.m. on a given day p(k<=0)=p(k=0)+p(k=1)=e ().

Let me tell you a little secret, the respondent is more interested in the number of admissibility than the apparent wealth value.

Because the reward for a certain number of admitives is much more than what you can give.

It is recommended that you only ask one question at a time. There will be a lot of people who will help you, so if you don't believe it, give it a try.

It is not easy to answer the question, please adopt it in time, thank you!

6 + 7 + 4 = 17, 17 choices in the first book x 16 choices in the second book = 272 choices, no order divided by 2 = 136 types.

a. The same subject: Mathematics 6x5 + Science 7x6 + Jingjing 4x3 = 84 kinds, no order divided by 2 = 42 kinds.

Number 1, number 2, number 1, number 3, number 1, number 4, number 1, number 5, number 1, number 6

Number 2, number 3, number 2, number 4, number 2, number 5, number 2, number 6

Number 3, number 4, number 3, number 5, number 3, number 6

Number 4 count 5 Number 4 count 6

Number 5 number 65 + 4 + 3 + 2 + 1 = 15, science and economics.

b. Different subjects: 6x7 in mathematics + 6x4 in mathematics + 7x4 in 7x4 = 94 kinds.

Numbers 1 1 Numbers 1 Subjects 2 Numbers 1 Subjects 3 Numbers 1 Subjects 4 Numbers 1 Subjects 5 Numbers 1 Subjects 6 Numbers 1 Subjects 7

Numbers 2 Subjects 1 Numbers 2 Subjects 2 Numbers 2 Subjects 3 Numbers 2 Subjects 4 Numbers 2 Subjects 5 Numbers 2 Subjects 6 Numbers 2 Subjects 7

Numbers 3 Subjects 1 Numbers 3 Subjects 2 Numbers 3 Subjects 3 Numbers 3 Subjects 4 Numbers 3 Subjects 5 Numbers 3 Subjects 6 Numbers 3 Subjects 7

Numbers 4 Subjects 1 Numbers 4 Subjects 2 Numbers 4 Subjects 3 Numbers 4 Subjects 4 Numbers 4 Subjects 5 Numbers 4 Subjects 6 Numbers 4 Subjects 7

Number 5 Subjects 1 Numbers 5 Subjects 2 Numbers 5 Subjects 3 Numbers 5 Subjects 4 Numbers 5 Subjects 5 Numbers 5 Subjects 6 Numbers 5 Subjects 7

Numbers 6 Subjects 1 Numbers 6 Subjects 2 Numbers 6 Subjects 3 Numbers 6 Subjects 4 Numbers 6 Subjects 5 Numbers 6 Subjects 6 Numbers 6 Subjects 7

6x7=42, the number of scriptures and the science of the scriptures.

Two independent chi-square distributions with degrees of freedom of 1 obey f(1,1). Therefore, it is necessary to try to organize the numerator and denominator into a chi-square distribution with a degree of freedom of 1, and organize it into a square form of a standard normal distribution. Molecular Part:

Normalizing x1+x2+x3 gives us a standard normal distribution, so take u=e(x1+x2+x3)=3. Then normalize the denominator.

After collation, it can be calculated that c=8 12=2 3

Solution: (1).According to the probability definition, the probability of the occurrence of the event should be f(x)x02, and the product should be 1

Accumulation 8c 3So c = 3 8

2).The product of f(x)x12 is equal to 1 2