Probability Permutations and Combinations Problem A probability problem related to permutations and

Let n=2k+1, then p(m=n) = c(2k,k) *1 2) (2k+1) *1 (k+1), where c(n,m) represents the number of different combinations of m in n numbers.

Finding c(2k,k) *1 2) (2k+1) is wrong because the solution is a binomial formula, and does not take into account the important constraint that m has always been on the right side of the x-axis until the last step.

This is a well-known problem in probability theory, called the bertrand voting problem'S Ballot Theorem), to the effect that two candidates, A and B, eventually receive P and Q votes respectively (let P>=Q) be the probability that A will not lag behind B in the number of votes during the roll-call. There are some information on the Internet that you can refer to, especially in English.

The landlord's question is the equivalent of the bertrand voting question. That is to say: in the process of random walking, the number of steps to the right is not less than the number of steps to the left, until the last step of the golden body is broken.

The move at the origin at 2k steps is c(2k,k), and we ask for the number of moves that are always "=0". The general idea is to fold, as shown in the picture above, if the gold body has not been protected before, the back of the way to move are all reversed, to the left to the right to go to the right, to the right to the left to go to the left... If the way is c(2k,k-1), then the way of walking without breaking the golden body is c(2k,k)-c(2k,k-1)=c(2k,k)*(1-k(k+1))=c(2k,k)*(1 (k+1)).

Solution: There are c(50,3)*c(47,3)*c(44,3)*c(41,3)*c(38,3)*....c(23,3) ways to install rivets, in which only one part is too weak, the number of types is c(10,1)*c(47,3)*c(44,3)*....c(23,3), the latter divides by the former, and the probability obtained is c(10,1) c(50,3)=1 1960

First choose a part, there are c(10,1) kinds, select 3 rivets a total of c(50,3) kinds, of which the strength is too weak only one, so the probability is c(10,1) c(50,3)=1 1960

I think the answer is questionable. I calculated that 3 19600 occurred that one component was too weak: 3 weak rivets were mounted on the same part and there were 3 options (which part was too weak), and the rivets on the remaining 2 parts had 3 out of 47 and 3 out of 44.

Total number of options: 3 out of 50, 3 out of 47 and 3 out of 44 for rivets on each of the 3 parts.

Probability p=3*c(3,47)*c(3,44) (c(3,50)c(3,47)c(3,44)))=3 19600

Embarrassment. Probabilistic theory. I just did it half a year ago.

To occur that the parts are too weak, first of all, the three rivets have to be selected.

c47 27/c50 30

Actually, the rivets are all in one component.

For: 10 cases.

Got out for 1 1960

But I remember that this is a practice problem after full probability, so the standard method is to use the full probability formula.

Randomly select 2 different gifts from 4 products, and there are a total of c(2,4)=6 types.

In the case of exactly one piece of the same variety, there is c(1,1)*c(1,3)=3 species, and the probability of exactly one piece of the same variety is 3 6 = 1 2

abcd

The chance of the second and first candidates for the same combination.

1/(c4 2)

The odds of 1 6 are exactly the same for both of them.

The first choice (C4 2) kinds, the second person chooses from the remaining two (C2 2) 681 = 6 cases, and none of the two people choose the same.

Total = 361 6 The odds are different for both of them.

Answer: It belongs to the classical generalization, each customer randomly selects 2 different gifts in 4 kinds of goods, there are c(4,2)=6 kinds, any two customers choose the situation of gifts, there are 6*6=36 kinds, there is exactly 1 piece of the same variety, c(4,1)*3*2=4*6=24 kinds, (first choose 1 piece of the same, the first customer chooses 1 piece, there are 3 methods, the second customer chooses 1 piece, there are 2 methods).

The probability is p=24 36=2 3

m departs from o, moves n steps, and reaches -1 point; The number of moving steps n=n, n must be odd, let n=2k-1, k is a positive integer, the total number of steps moved to the right is k-1, and the total number of steps moved to the left is k; p{n=n}=[(1 2) (k-1)][1 2) k], so p{n=n}=1 2 n.

1. If you don't put it back, every time you draw it, it will affect the result of the next time, but it's not like you said that you want to use it, because what his title requires is: the first white ball, the second time you get the red ball, that is, you must drop in this order, so you can't use A to arrange; 3. In the question: Twice to get a red ball and a white ball, it may be the first red ball and the second white ball; It could also be the first time you get a white ball and the second time you get a red ball.

Therefore, it is necessary to use A22 drops...

1. The first question of the first question has been divided into the order, there is no need to use A, the second question is not divided to use A, and the second question is not divided to use A, so whether we need to use A, we must depend on the situation of the question: if there is no score, we must use A, and if you score, use the first question of the first question: because the question requires the probability of taking the white ball for the first time and taking the red ball for the second time, the order has been divided, and we don't need to divide it again.

We have to follow this order (the first time we have to draw a white ball, the second time we have to draw a red ball), so the first time we take one white ball from two white balls, and the second time we have to take one red ball from five red balls, although there is no requirement to do so, there are two kinds of results to obtain, so we must use the permutation. A means arrangement, since it is an arrangement, it should be in order, two times to take a red and a white, there are two possibilities: the first time to take the red ball, the second time to take the white ball:

The first time you take the white ball, the second time you take the red ball, so you need to use the two results to divide the order of success. Work hard, work hard to give points! Thank you ha!