Probability questions that need to be determined, questions about probability

Only the above answer of 2 9 is correct, but the solution is too complicated.

Reducing the round table to a queue, assuming that the husband does not participate in the queue, the remaining 9 people are lined up and the wife is seated next to each other when she is first or last.

Assuming n people line up, the probability of a particular person coming first is 1 n, and the probability of being last in line is also 1 n, which adds up to 2 n. In addition to the husband, there are 9 people in line, and the wife has a chance of being first or last 2 9. After simplifying the model of the round table to a queue, I know that when the wife is first or last, the couple is next to each other.

So, in the end, the correct answer is 2 9.

If there are 3 people (including the couple) and n is 2, the probability of the couple sitting next to each other is 2 2 = 1.

That's it. The key to solving the problem is the simplification of the mathematical model, otherwise it will be dizzy.

First of all, the total public has a ranking of 10! = 3,628,800 species.

Secondly, there is no left and right seat for the husband and wife, as long as they are next to each other, so it can be set as a husband and wife as a whole, that is, it is counted as 9 people, but in this case, the result will be multiplied by 2 (because the husband and wife can be in different orders).

n=9！*2=725,760 species.

Total possible case p10 (10) = 3,628,800

Choose a seat for one of the couple, there are 10 options, and the spouse's seat has only 2 options: 10*2=20

Probability bit 20 3268800 = 1 1634400

This is the placeholder problem, counting the husband and wife as one, then there are 8 people left after the husband and wife, they sit around the round table, there will be an empty seat between the two people, a total of 8 empty seats, as long as the couple is arbitrarily placed in one of the empty seats to sit together, so the probability that the couple is sitting together is 1 8

Method A (10,10) to sit casually

A round table selects 2 hungry enclosures connected by the method of C10 to choose 1 The couple's seat can be changed and then multiplied by 2, and the remaining 8 seats are arranged at will A (8, 8).

The probability is! 2/9

I hate probability the most. Why is it so complicated? Is it on a 50% probability! Together or not!

Categories: Education, Science, >> Learning Aid.

Problem description: In the probability operation, the symbol c or a will be used, as if one is in order and one is not in order, I would like to ask about their difference, thank you in advance.

Analysis: A is sequential, C is unordered.

To put it simply, it is the same number, and a is larger than c.

For example, Qi Baofeng said that there are three ABCs, and there are several ways to select two people.

It should be calculated with c because there is no order, i.e., ab ac bc (c32=3*2*1 2*1).

It can be seen that AB and BA are actually the same.

And if it is said that Gao Sheng selected 2 people to do the work of A and B respectively, there is an order.

For ab means A as A and B as B, and ba as B as A.

So use a to calculate (a32=3*2*1=6).

That is, there are six kinds of ab ba ac ca cb bc.

p(a∪b∪c)=p(a)+p(b)+p(c)-p(ab)-p(ac)-p(bc)+p(abc)=7/8

So p(abc)=p(a, b, c)-(7 8)=1-(7 8)=1 8

If events a, b, and c are independent of each other, then p(abc) = p(a)p(b)p(c).

If events a, b, and c are not independent of each other, i.e., whether event a occurs or not is related to event b or event c, then p(abc) is not equal to p(a)p(b)p(c).

(1) Classification discussion.

Suppose you choose a class because there is a sequence, then use the permutation probability as c101 c491 a502 1 5

In the same way, it can be obtained: Suppose you choose the second class, the probability is c181, c291, a302, 3, 5, the probability is added to 4, 5

2) It's basically the same as the first question.

I'm sorry, I can't seem to play the upper and lower corner markers, I hope you understand C101 10 is below 1 is above

It should all be 7 20 (1) 10 + 18 50 + 30 = 7 20

2) It has nothing to do with the girls selected first, because it should be the same method 7 20

I'll just assume you're asking for the second equal sign.

The parenthesis in front of it is equal to b, and the second is equal to b. A is not B and AB is equal to B, and it is the common part, so the part they have in common is B, and A and A are not intersecting space-time sets. The next bracket is the same.

As for the mathematical proof, I remember that in probability theory, it is not the multiplicative distributive law that is true, then you can treat a and b as numbers and finish it, for example, the previous one is (a non + b) (a + b), get (a non + b) a + (a non + b) b, and then get aa non + ab + a non b + bb, which is not equal to b. The middle bracket at the back is based on the symmetry of the text, and directly replace b with b non, which together is bb non.

I won't, but I can't help but want to compliment you, your handwriting is so beautiful.

x line lead book 1000 = 6 (1-p 30) = p = 30 - x 200

Profit Sales Cost p*x - c = 30 - x 200)*x - 5x - 25000 = x 2) 200 + 25x - 25000 = 1 200*(x-2500) 2 + 7250 <=7250

The maximum profit of the agitated macro is 7250, where x=2500=>p=30 - 2500 200=

1. Any two of the six pants chaotic cherry blossom scientists have a total of 15 bell combinations, including the combination of A by 5 bells, so the probability that the ** scientist A will accompany Lu to be selected.

p=5/15=1/3

2. A **scientist and a female scientist, but Hu Cong** scientist A and female scientist A cannot be selected at the same time, there are 3 3-1 = 8 combinations. Then the probability of selecting a ** scientist and a female scientist:

p=8/15

1 There are six people in total, and if two people are chosen, the probability is one-third of the scum.

2 If A and A are not selected at the same time, there are 14 cases (5+4+3+2+1-1), and if a male beam makes a female combination, there are 8 types, so the probability of Lu liquid is 4 out of 7. Forget.