Let basketball team a and team B play. There is one winning team in each game. If one team wins four games, the game ends. Assuming that the probability of a and B winning in each game is half, try to find the expectation of the number of matches If X My calculation: P (x = 6) = C (2,5) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * 2 = 5 / 8), is there any error in this?

Let basketball team a and team B play. There is one winning team in each game. If one team wins four games, the game ends. Assuming that the probability of a and B winning in each game is half, try to find the expectation of the number of matches If X My calculation: P (x = 6) = C (2,5) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * (1 / 2) * 2 = 5 / 8), is there any error in this?

One half less
X = 6, how can it be 5 1 / 2

A simple mathematical problem of permutation and combination Six people line up. A doesn't stand at the head or the end What I do is C (4,1) * a (5,5). Is the front C (4,1) right? I want to choose a position from the middle to let station A. another solution is to subtract the head and tail of a row from a (6,6). It seems that the reduction is more. I am not very clear about what has been reduced. Please help me answer, thank you

C (4,1) * a (5,5) is right
A (6,6) minus the head and tail of a row, i.e. minus C (2,1) * a (5,5)

A mathematical permutation and combination problem Among the 200 products, there are 2 defective products, and 5 of them are selected randomly (1) How many kinds of sampling methods are there for "just two defective products"? (2) How many kinds of "one of them is defective"? (3) How many kinds of "no substandard products" are drawn? (4) How many kinds of sampling methods are there for "at least one defective product"? Please write the formula (e.g. 5c2 for the combination of two out of five, and 5A2 for the arrangement) thank you!

Solution: 2 out of 200 products are defective and 198 are genuine
(1)(2C2)(188C3)
(2)(2C1)(188C4)
(3)188C5
(4)(2C2)(188C3)+(2C1)(188C4)
Simplify and finish it by yourself. Remind you to read more textbooks

On a mathematical problem of permutation and combination Take a simple example as an example At present, there are 3 persons from Party A, Party B and Party C. among them, 2 people are selected to attend an activity. What is the probability of a being selected? I have two ways to solve problems (because I didn't go to school, the arrangement and combination in the high school textbook is self-study, so I don't know the standard.) The first method is to solve the problem by permutation a. First, take 2 out of 3. There are six permutations, that is, the number of permutations is a = 6, b. A can be divided into two categories: ① a, X: 1x2 ② X, a: 1x2, c. Then the probability of a being selected is: (1 + 2): a ☞ 2 / 3 The second is to solve the problem by combination a. Similarly, if you take 2 out of 3, there are three combinations, that is, the combination number b = 3, b. Because it is composed of two people, a occupies a position after being selected, and the remaining position can be B or C, So the combination number C: 2 takes 1, there are two kinds, that is, C = 2, c. Then the probability of a being selected is: C: B = 2 / 3 doubt: 1. Is the above solution correct? 2. Is there anything wrong, or is there an inappropriate description? 3. If it's correct, I'll ask another question to see if you can use these two ideas to solve this problem. No, it's to extend this idea to other topics There are 15 small balls of the same size and shape, including 6 white balls, 5 black balls and 4 red balls. If three balls are taken from the same three directions, what is the probability of getting one white ball, one black ball and one red ball? 4. Answer with the above two ideas? 5. Is there a better idea?

Definition of arrangement:
Generally, m (m ≤ n) elements are taken from n different elements and arranged in a column in a certain order. According to the definition of permutation, the two permutations are the same if and only if the elements of the two permutations are identical and the order of elements is the same. For example, the elements of ABC and abd are not identical, For example, ABC and ACB, although the elements are identical, the order of elements is different, and they are also different
Definition of combination:
Taking n elements at a time from m different elements, no matter in what order and forming a group, is called combination
For example, 231 and 213 are two permutations, and the sum of 2 + 3 + 1 and 2 + 1 + 3 is a combination
Now to answer your question:
The above solution is correct
However, if you also use the above method 1 (i.e. using the method of arrangement), the process is very complicated. It requires that all possible permutation orders are listed before the probability can be calculated
a. First, take three out of 15, so a = 15 * 14 * 13 = 2730
b. If you take one white, black and red ball, you need to take one out of six white balls, five black balls and four red balls. There are 6 * 5 * 4 = 120 kinds of methods to take out, and then arrange the balls taken out, B = 120 * 6 = 720 different arrangements
c. So the probability is: B: a = 24 / 91
Obviously, it is very convenient to use the combination method here. Because we don't consider the order of the color of each ball, we directly use the combination method
a. First, take 3 out of 15, so B = (15 * 14 * 13) / (3 * 2 * 1) = 455
b. Since one is required to be taken from each white, black and red ball, C = 6 * 5 * 4 = 120
c. The answer is C: B = 24 / 91
A better way is to use the method of distribution function in University probability statistics, which can be solved directly by using hypergeometric distribution formula

The permutation and combination problem in senior high school mathematics, how to distinguish the problem of insert space, partition board and pile separation

1. Insert space is used to solve the problem of non contiguity. For example, if a and B can't be adjacent to each other, then take 4 people except a and B to arrange first, and then take a and B to insert space, because a and B have different space, so they must not be adjacent. 2. Partition method is used to group and get the element serial number of multiple elements

Please elaborate the "partition method" in permutation and combination problem

The partition method requires the undifferentiated balls to be divided into ordered piles
Since there is no difference between the "balls", the number of each pile can only be reflected, but not which ball it is
1. Empty heap is not allowed
Example: the positive integer solution of X + y + Z = 10
Put two boards in the air to form three parts
2. Empty heap is allowed
Example: the nonnegative integer solution of X + y + Z = 10
10 "balls" and two boards occupy 12 positions to find two places to place the plate

The application of the method of permutation and combination clapboard! There are 15 balls put into four boxes numbered 1, 2, 3 and 4. The number of balls in each box is not less than its number. How many different ways are there?

First put 0,1,2,3 balls into four boxes numbered 1,2,3 and 4
So just put at least one more ball in each box
The remaining nine balls are arranged in a row, and three separators are inserted into the eight spaces in the middle, and the nine balls are divided into three piles. The clapboards can not be adjacent. Therefore, there are 56 kinds of release methods for balls, C (8,3) = 56 (C is the combination number)

The application of high and middle school permutation and combination partition method There is a question that is very puzzled: how to divide 12 identical balls into three different boxes, with at least one in each box, and the method of dividing 12 identical balls into three identical boxes with at least one in each box. How are the answers to the two questions different

The answers of the two questions are completely different. The former one has no problems with the partition method,
Compared with the first method, the second method should be discussed by classification
1. If the quantity in three boxes is the same, there is one method (in the first method, this case also counts one method)
2. There are only four methods for the same quantity in two boxes (the same quantity can be 1.2.3.5) (in the first method, three methods are calculated for each case)
3. There are 7 kinds of the same quantity in any box (1 + 2 + 9 1 + 3 + 8 1 + 4 + 7 1 + 5 + 6 2 + 3 + 7 2 + 4 + 6 and 3 + 4 + 5)
(in the first question, six cases are counted in each case)
Therefore, there are only 12 methods for the second method. The first method has 11 * 10 / 2 = 55 methods

Why do we need to solve problems with the method of inserting boards? Put 10 pieces of sugar into 3 baskets, why not use C10 (bottom) 2 (top)

If we use the method of inserting boards, we will not mix them up, we will not calculate more or less

Mathematical factorial operation (n!) ^ 2 / (2n)! How to calculate it? I'm always n + 1, but the answer is (n + 1) ^ 2 / [(2n + 2) (2n + 1)],

Obviously (2n)! = n! * (n + 1) (n + 2) (n + 3). 2n
And (n!) ^ 2 = n! * 1 * 2 * 3 * 4
So (n!) ^ 2 / (2n)! = 1 * 2 * 3 * 4. N / (n + 1) (n + 2) (n + 3). 2n
The limit is 0, so it is convergent