The question about high school probability is very urgent!

The probability of taking out 0 red balls from box A is c3 to take 2 C4 to take 2 = 1 2, and the probability of taking out 1 from box A is 1-1 2 = 1 2

The probability of taking out 0 red balls from box B is c4 to take 2 c6 to take 2 = 2 5, the probability of taking out 2 is 1 c62 = 1 15, and the probability of taking out 1 is 1-2 5-1 15 = 8 15

1) The probability of exactly 1 red ball out of the 4 balls taken out = the probability of taking out 0 red balls from box A * the probability of taking out 1 red ball from box B + the probability of taking out 1 red ball from box A * the probability of taking out 0 red balls from box B = 1 2 * 8 15 + 1 2 * 2 5 = 7 15

2) What is the probability of a red ball in the 4 balls taken out? It's not clear how to ask the question, is it the probability of taking out a red ball in 4 balls, or the expectation of taking out a red ball?

The probability that there are no red balls in the 4 balls taken out = the probability of taking out 0 red balls from box A * the probability of taking out 0 red balls from box B = 1 2 * 2 5 = 1 5, so the probability of having red balls is 1-1 5 = 4 5

The probability of 2 red balls out of 4 balls taken = 1-7 15-1 5-1 30 = 3 10

The expectation of the red ball in the 4 balls taken out = 0 * 1 5 + 1 * 7 15 + 2 * 3 10 + 3 * 1 30 = 7 6

Make a list of all the possibilities and add them up at the end, 1 4 4 6 3 5 3 4 1 3 4 6 3 5 3 4 2 3 2 6 4 5 3 4 2 3 4 6 2 5 I won't write the second question, I'm too tired....

Option D solution 1: A and B meet, there are no more than two possibilities:

1) Same group. 2) The probability of winning (1) in different groups, but in the group stage (in fact, the "semi-finals"): because in addition to A, there are 3 teams, the probability of A and B in the same group is 1 3(2) (i) The probability of different groups is 2 3, ii) each defeats the opponent in the same group, and the probability of meeting the division in the final is (1 2) * (1 2) = 1 4 (multiplication principle, evenly matched).

Therefore, the probability that different groups of A and B still meet is: (2 3) * (1 4) = 1 6 (1) + (2): the probability of meeting is (1 3) + (1 6) = 1 2 Solution 2:

All conditions are symmetrical (evenly matched, randomly grouped), and each team encounters an average of one opponent (1 game and 2 games for group losses and wins, (1+2) 2=, and according to the format, each opponent is a different opponent (the team that plays another group in the final).

There are 3 opponents in total, so the probability of encountering a specific opponent is:

If A and B are grouped together, then they must meet, and the probability at this time is 1 6

If A and B are not in one group, then the probability of the two meeting is 1 2 * 3 1 * 2 1 = 1 12 (1 3 is the probability that A and B are not in one group, and the two 1 2 are the probability that A and B will win in their respective groups).

The sum of the two is 1 4, choose B

There are two scenarios: First: The probability of meeting in the first game is:

Draw out a random person first If it is D or C, then it must meet C or D If it is A or B, then it must meet B or A So the probability is 1 3 Second: The probability of meeting in the second game is: Draw a random person If it is A, then it must be against C or D The probability is 2 3*1 2*1 2=1 6 The first and second two cases are added together = 1 2 So choose D

d See solution 1 of apsifio. As a multiple-choice question, after calculating the probability of the same group as 1 3, you will know that the result must be greater than 1 3. Solution 2 is profound, I don't understand.

There are three types of grouping: (1) A and B; Propylene buty.

2) A C; Ethyl D.

3) Methyl Ding; Ethyl-C.

The probability of A and B meeting in groups is 1 3;

In the remaining two cases, the probability of A and B meeting is 1 2*1 2*1 3=1 12, a total of 1 12*2=1 6;

So 1 3 + 1 6 = 1 2

A1 2

Solution: Let Xiao Ming be A, and the four friends are BCDE. The winner is credited with (x, y).

Then all the results of the winning lottery are: (a,b)(a,c)(a,d)(a,e)(b,c)(b,d)(b,e)(c,d)(c,e)(d,e) a total of 10 kinds.

If "one of the selected events is Xiao Ming" is Q, then there are four types of event Q: (ab) (ac) (ad) (ae).

So p(q)=4 10=2 5 so the probability is.

Either you got the question wrong, or you got the answer wrong.

Total number of combinations: c (4 out of 10) = 210

The number of combinations containing A and B at the same time: c(8,2)=28

Number of combinations that are all female: 1

Number of 3 non-B female combinations: 1*c(6,1)=6

Number of combinations of 3 women including B: c(3,2)*c(5,1)=3*5=152 number of combinations of non-B women: c(3,2)*c(6,2)=3*15=452 combinations of women including B:

c(3,1)*c(5,2)=3*10=30, then the probability of meeting the requirements is: (1+6+15+45+30) (210-28)=97 182

If 2 male and 2 female representatives are elected:

First select 2 male representatives from 6 male students, there are c(6,2)=15 options, and then select 2 female representatives from the remaining 4 female students, and there are c(4,2)=6 selectionsIf 4 female representatives are elected:

4 female representatives were selected from 4 female students, and there were c(4,4)=1 selections, and a total of c(6,2)*c(4,2)+c(4,4)=91 selections.

There are two categories:

Contains B: (C3,1C5,2+C3,2C5,1+C4,4) (C10,4-C8,2).

Excludes B: (C3,2C6,2+C3,3C6,1) (C10,4-C8,2).

Add it up and sum it up.

All basic events are all the different permutations of the four colors, a total of a(4,4) = 24.

The probability of the fourth person drawing the black color pen is 1 4, in fact, no matter how many positions are drawn, the probability is 1 4, which means that the order of the draw does not affect the fairness. The reason is simple, the favorable event for the 4th person to draw the black is: the 1st, 2nd, and 3rd people did not draw the black pen.

There are 3 * 2 * 1 = 6 kinds of this event (the first person draws 1 of the 3 sticks except for black, there are 3 kinds, the 2nd person can only draw 2 black sticks, and the 3rd person can only draw 1 stick except for black), so the probability of the 4th person drawing black is 6 24 = 1 4

The following proves that the probability of each person drawing black is 1 4: (which can be used as a typical example of understanding conditional probability).

1. The first person has four choices, and the probability of drawing is 1 4

2. There are two situations when the second person draws: if the first person draws black (probability is 1 4), the probability of the second person drawing black is 0, and if the first person does not draw black (probability is 3 4), then the second person has 3 choices, and the probability of winning black is 1 3, so according to the conditional probability, the probability of winning is 0 * (1 4) + (3 4) * (1 3).

3. The third person can be obtained according to 1-1 4-1 4-1 4, or it can be calculated directly according to the conditional probability, which is similar to the calculation of the second person.

"When the red dice are 4 or 6" means that the red dice are either 4 or 6 (probability of 1 2 respectively). If it is 4, then the product of the two is greater than 20, and the yellow dice can only be 6, (the probability of occurrence is 1 6); If it is 6, then the product of the two is greater than 20, and the yellow dice may be (the probability of occurrence is 3 6 = 1 2). So, the total probability = 1 2 * 1 6 + 1 2 * 1 2 = 1 3, choose b.

(1) When red is 4, yellow is 6, and the probability is 1 2 *1 6(2) When red is 6, yellow is 4, 5, and 6 probability is 1 2 *1 2 The two cases are added to get 1 3

I think the first question lz is right, the title only says 2 poles, which is the default wire length is 100 meters, that is, the 100 meters between the two poles, the damage range is 10 meters plus 10 meters is 20 meters, the probability is 1 5, if the 10 meters outside the two poles are counted, it is equivalent to saying that there are wires outside the two poles, but this situation is not said in the title, according to the reality of the length outside the pole tends to be infinite, the probability tends to be close to 0.

If your math teacher is very responsible for this assignment, it is recommended to talk to your teacher, the question is obviously flawed and the answer is not realistic.

The second problem can be solved quickly by drawing a picture.,I'm not going to say much about the second question after I've said it upstairs.,First of all, in this plane coordinate system, (0,0)(0,1)(1,1)(1,0)The X value and Y value corresponding to each point in the rectangle enclosed by these 4 points correspond to 2 randomly selected 1 number (these 2 times are sequential), then this area is the total occurrence (how to say this term, I forgot), and then the next is to select the point where the two numbers are added to just that, Because the line practiced by this point is the dividing line that divides the probability inside and outside the probability, by selecting a few special points such as (, 1) (, , etc., you can find that this line is a straight line through (1, and (, 1), and then calculate the probability that the area of this small triangle is the area of the upper block. It's a hassle to say, but it shouldn't be that complicated when you actually do it.

Upstairs answered correctly.

It's just that the second question should be [1-(.]

=2, that is, when the touch is red-yellow or yellow-red. Then p1 = 3 5 * 2 4 + 2 5 * 3 4 = 3 5 =

At 3 o'clock, that is, when the ball is touched is red-red yellow or yellow-yellow-red. Then p2 = 3 5 * 2 4 * 2 3 + 2 5 * 1 4 * 3 = 3 10 =

4, that is, the touch of the ball is red-red-red-yellow, then p3=3 5*2 4*1 3*2 2=1 10=

Just write down the distribution column yourself.

All possible values are = 2,3,4

2, the first time to touch the yellow ball, the second time to touch the red ball or the first time to touch the red ball, the second time to touch the yellow ball = 3, the first two times to touch the yellow ball, the third time to touch the red ball or the first two times to touch the red ball, the third time to touch the yellow ball = 4, the first three times to touch the red ball, the fourth time to touch the yellow ball.

= 2, 1 out of 3 out of 1 out of 1 * 2 out of 1 out of 4 out of 1 + 2 out of 1 out of 5 out of 1 * 1 out of 1 out of 4 = 2 out of 5

3, 1 out of 3 out of 5 out of 1 * 2 out of 1 out of 4 out of 1 * 2 out of 1 out of 3 out of 1 + 2 out of 1 out of 5 out of 1 * 3 out of 1 out of 1 out of 1 * 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out of 1 out