Acute high school three probability problem, high school three probability problem

Thinking about 6 times 4 times, each customer can only choose one window, so there are 6 possibilities for the first customer to choose the window, and then the other customers in turn can be found as 6 4 times through the principle of multiplication. Any mathematical problem can be solved from different angles, and then we think backwards and choose customers for the window: a window may have 0-4 customers, which needs to be classified and discussed, which can be expected to be troublesome.

Therefore, we can conclude that such a problem can be calculated from an object that can only have one selection ability, for example, in example 1, select a window for customers, select an interest group for students in 2, and select a mailbox for a letter (although it is not said in the question that a letter can only enter one mailbox, we can determine it through common sense). For the last question, I hope you can take the original question so that you can analyze and solve it according to the specific situation.

This kind of problem cannot be memorized. ( indicates several parties).

1, 4 customers each choose a window, then each person has 6 choices, then.

2. The same goes for it. There are 3 choices for each person, then 3*3*3=3 3=273, the same goes for them. There are 4 choices for each person, then 4*4*4*4*4=4 54, a5(5)-[a4(4)*2-a3(3)]=78 is arranged first, which is 5!

A does not perform the first job, B does not perform the second job, a4(4)*2, double count, minus 3!Got 78

The first 6*6*6*6

Think of it this way, everyone has their own choice, there are four people in total, six types each.

Then the following 3*3*3*3

Question 3 4*4*4*4*4

You can't think that it's the mailbox that chooses the letter, and it should be the letter that chooses the tube.

1.Each person has 6 choices, a total of 4 people, which is 4 times of 6.

The following are similar.

There are 40 of them.

Let there be a in b, then the probability of drawing A and B in b is 2 times 1 a times 1 (a-1) = 1 28

Dismiss a = 8, so there are 8 in b.

So there are 40 of them.

Because the probability of at least one volunteer in class A in the logistics group is 37 42, so none of the volunteers in the logistics group have destroyed the eggplant, and the probability of some is 5 42.

1-(x*(x-1)*(x-2) (9*8*7)=5 42 so x=4 so there are 5 people in class B.

p(ε=3)=1/21

p(ε=2)=4/21

p( =1)=10 Sidver21

p(ε=0)=2/7=6/21

The distribution columns are as follows.

p 1 21 4 21 10 21 2 7e = 0*1 21+1*4 21+2*10 21+3*2 7=2 The intermediate probability calculation is too troublesome, so I didn't type, if you don't know, send me a message.

LS does a good job of correcting the exigent residue and the final distribution of the demo column is.

p 1 21 4 21 10 21 2 Ming liquid 7

eε=1

There are several scenarios; 1: 2 people, 2 people. 2: 1 person, 3 people. 3: 4 people. 4: 2 people, 1 person, 1 person.

The first: 3 6 5. The second: 4 6 5. The third: 6. Fourth: 6 6 5 4.

The overall case is 6 4 with a probability of 13 18

Roll the dice twice in a row to get the number m,n of the mountain guess type, and the angle between the vector comic (m,n) and the vector (-1,1) is greater than 90 degrees of mega leakage.

then m>n

m=6,n=1,2,3,4,5

m=5,n=1,2,3,4

m=4,n=1,2,3

m=3,n=1,2

m=2, n=1, a total of 5+4+3+2+1=15 situations meet the requirements.

The probability is 15 6 * 6 = 5 12

Before you learn the combination, let's use the simplest probability first.

1) (3x2x1x3+3x2x1x6) (6x5x4) = 9 201 first class, 2 second class or 1 first class, 1 second class, 1 third class.

You should know if you multiply 3 and 6 respectively, right? There are several different orders when drawing)2)(3x2x2x3+3x2x1x3) (6x5x4)=9 202 first class, 1 second class or 2 first class, 1 third class.

3）(3x2x1+3x2x2x3+3x2x1x3)/(6x5x4)=1/2

3 first class or 2 first class, 1 second class or 1 first class, 2 second class.

If you don't understand, ask again.

1) There is exactly 1 first-class probability = c3 (1) * c3 (2) c6 (3) = 9 20

2) There are exactly 2 first-class probabilities = c3 (2) * c3 (1) c6 (3) = 9 20

3) No third-class probability = c5 (3) c6 (3) = 1 2

For set a, z=[1-i (n+1)] (1-i)=[1-i (n+1)](1+i) 2 (n n).

According to the exponent of i, there is i n (n n) so a==

And b = so the possible values of w are 0, 1, i 1, i, 2i, i, i, i, i, i, i, i, i, i, Pick D

The set of a is 1,1+i,i,0

The basic events of the b set are 4*4=16, and the ones that meet the requirements are z1=1 z2=1+i

z1=1+i z2=1

z1=1+i z2=i

z1=i z2=1+i

So choose b for 4 16 = 1 4

a=b=

1+i, the root of i-1 is root number 2

So choose the reciprocity of the d-superior set.

i should be an imaginary symbol, when n=4n+1, the z module in a is the root number two, and when n=2n, the z module in a is 1, so the probability of the two is 1 4 and 1 2 respectively, the element modulus in b is the product of the modulus of any two elements in a, and the event must have two element modulus values of the root number two and 1, and its probability is 2*1 4*1 2=1 4, choose b