Senior sophomore math probability, math probability sophomore sophomore

The quantity produced: 1 2 + 1 3 + 1 6

The number of defects: 1 2* + 1 6*, the probability of defects: the number of defects divided by the quantity produced (you can calculate it yourself).

Number of defects for A1: 1 2*

The series of production A1: 1 2

Probability of a1: Divide the two.

Probability of defective product = 1

Probability of production by a1 = (1

1. Regardless of the constraints, the total arrangement = 5 4 3 2 1 = 1202, the arrangement of A at the left end and adjacent to B = 3 2 1 = 63, the arrangement of A at the left end but not adjacent to B = 3 3 2 1 = 184, A is not at the left end, and A is on the left side of B = 3 2 1 3 = 185, A is not at the left end, and A is on the right side of B = 3 2 1 4 = 246, answer = 120-6-18-18-24 = 54

Master programme a(5)(5).

A at the left end of A(4)(4).

A and B adjacent to 2a(4)(4).

Treat A and B as a person, and you can transpose them).

A(3)(3) where A is at the left end and adjacent to B

Only the remaining three people are arranged)

So a(5)(5)-3a(4)(4)+a(3)(3).

Assuming that the total output is 300, it is very simple to divide it among three companies!

1.There are 5 in total! Assuming that ball 3 is in box 3, there are 4 balls left! arrangement. If so, the probability is 4! /5！=1/5

2.This is equal to 1 (all possibilities) - (probability of ball 1 in box 1) - (probability of 5 balls in box 5) + (probability of 1 ball of 5 at the same time in box 1 5) = 1-1 5-1 5 + 3! /5!

3.Here is a formula dx=put x balls into x positions, so that all the finds are not in their own positions (i.e. number 1 is not in the first place, and so on) is x!(1-1/1+1/2!

1/3!+.All the way up to x).

So the distribution is:

0： d5/5! =5!(1/2!-1/3!+1/4!-1/5!)/5!= 11/30

1: d4/5! =5! (1/2!-1/3!+1/4!)/5!= 3/8

2: d3/5! =5!(1/2!-1/3!)/5!= 1/3

3: d2/5! =5!/2!/5!= 1/4

4: d1/5! =0

5: d0/5! =1/5! =1/ 120

So the expected value = 0*11 30+1*3 8+ 2*1 3 + 3*1 4 + 4*0 + 5*1 120 = 44 24= 11 6

That's it, the answer is uncertain (because there's no paper around), but the concern is right.

Of the 4 keys, only one is an office key.

4 keys are opened after one attempt The probability is p1=1 44 keys are opened after two attempts The probability is p2=3 4x1 4=3 16 The probability of opening the door no more than twice is p=p1+p2=1 4+3 16=7 16

The probability of opening at one time is 1 4

The probability of two openings is 3 4 * 1 3 = 1 4

Add 1 2 twice

The probability of opening a door at one time is 1 4

The probability of opening the door twice is 3 4 * 1 3 (the first time you get it wrong, the second time you get it right) = 1 4

So the probability of opening the door no more than twice = 1 4 + 1 4 = 1 2

The combination symbol is not written, you see my handwritten solution (picture)!

1. The number on the three-digit number is 1,4,7 or 2,5,8, and there are 2a(3,3) = 12;

2. The three-digit number contains only one of 0,3,6,9, and the number on the other two digits is taken from (1,4,7) and (2,5,8) each one, such numbers have c(4,1)c(3,1)c(3,1)a( but to remove the number of 0 in the hundred, there are c(3,1)c(3,1)a(2,2), so there are 216-18=198;

3. The numbers on the three-digit numbers are 0,3,6,9 in 3, but to remove 0 in the hundreds, so that there should be a(4,3)-a(3,2)=18, in short, from 0 to 9, these 10 numbers constitute no duplicate numbers and can be divisible by 3 3 digits have 12 + 198 + 18 = 228, the first combination of not 0 is a(10,3)-a(9,2) = 648, then the three digits that are not divisible by 3 are 648-228 =420, so the probability is 420 648=35 54

Therefore, that answer is wrong.

15) There is a high probability of encountering the opposite sex.

Solution: The probability of meeting the same sex is 24 49, and the probability of meeting the opposite sex is 25 49 The probability of meeting the opposite sex is high (this question discusses the uniform probability situation, not the situation similar to the high probability of meeting a man in a sports store and the high probability of a woman in a clothing store).

Solution: The first time you draw a card, any suit is OK, and the second time you draw a card of the same suit as the first time is 13 52 = 1 4

Solution: The chosen combination of A is (A, B), (A, C), (A, D), (A, E) All combinations are c(5,2)=10

The probability is 4 10 = 2 5

There is a high probability of meeting classmates of the opposite sex.

15: Regardless of whether this classmate is male or female, then he only has 24 people of the same sex and 25 people of the opposite sex, so the probability of meeting the same sex is 24 49, and the probability of the opposite sex is 25 49.

9: Whether it is the same color before and after the two draws, it depends on the operation of the second time, and it has nothing to do with what suit it is drawn out for the first time, so the probability is 13 52, that is, 1 4

12: The combination with armor is c(1,4), the total combination is c(2,5), the probability is c(1,4), and c(2,5) is 2 5.

Hope it helps.

There are a total of 4*4*4*4 types of coating methods.

Suppose 2 has 4 kinds, then the adjacent 3 has 3 kinds, and now 4 is divided into two cases:

When 4 and 2 are painted the same color, then there are three kinds of 1, i.e., 4*3*1*3;

When 4 is different from 2, there are 2 kinds, and now 1 has 2 kinds, that is, 4*3*2*2 The result is: the same divided by 4, :(3*3*1+3*2*2) 4*4*4=21 64

The binomial distribution is the distribution of the number of events in n experiments, and it can be said that each independent experiment is a two-point distribution. A Shelley experiment is equivalent to a two-point distribution, and this relationship is very important, and the mathematical expectation and variance of the binomial distribution can be deduced from this idea.

As for the geometric distribution, the example problems in the book are on the line, and they are not commonly used.