A middle school math competition question 20 on probability

5 on one side, 2 on one side: 6 cases on one side of 5 (blank), 36 cases on one side of 2 (this is simpler), a total of 6x36 = 216 types.

4 on one side, 3 on one side: there are 35 on one side of 4, 33 on one side of 3, (both of which I did on the basis of interjection, I didn't think of a good way) a total of 35x33 = 1155

216 + 1155) x2 = 2742 species.

If they are not adjacent, you can abstract them into 10 numbers to find out how many cases there are when two numbers are not adjacent. According to the drawer principle, if you find 6 numbers at random, there must be adjacent numbers, so there are up to 5 people on one side. The two sides of the corridor are A and B, and when there are 5 in the corridor, there are 1 3 5 7 9 , 2 4 6 8 10.

1 4 6 8 10 , three cases. B is equivalent to taking two of the 10 numbers and subtracting the adjacent cases, a total of 36 kinds. Total 36 * 3 = 108 species.

There are also cases where there are 4 people on side A and 3 people on side B, and A has a total of them. 1 3 5 7 , 1 4 6 8 ,1 5 7 9 ,1 6 8 10 ,2 4 6 8 , 2 5 7 9 ,2 6 8 10 ,3 5 7 9 , 3 6 8 10, 4 6 8 10 , a total of 10 kinds, B side is equivalent to 3 numbers Not adjacent, we analyze 1When the first and third rooms are occupied, there are 6 types in total.

When the first and fourth rooms are occupied, there are 5 types in total. When the first room and the fifth room are occupied, there are 4 types in total, and so on, for a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 types. 2.

When the second room and the fourth room are occupied, there are 5 types, and so on, there are 5 + 4 + 3 + 2 + 1 = 15 types, 3When the third and fifth rooms are occupied, there are 4 types, and so on 4+3+2+1=10. 4.

There are 6 types of 4th and 6th rooms when occupied, 5When the 5th and 7th rooms are occupied, there are 2+1=3 types, 6There is one type of room in the 6th and 8th rooms when they are occupied.

In total, there are 56 species on side B, and 560 species on A and B. There are 560 kinds when there are three people on the first side, and 108 kinds when there are two people on the first side, totaling 108 + 560 + 560 + 108 = 1336 types.

Ask for all the possibilities (no conditions): each vertex has 4 possibilities (0 red, 0 blue, 1 red, blue), so there are 4 4 4.

Then find the opposite event, that is, the number and color of any two adjacent vertices are not the same, the first vertex has 4 possibilities, the two vertices adjacent to it have one can, and the one opposite it has one possibility, so there are 4 possibilities in total.

So there is a total of 4 4-4 = 252

For example, when considering ABC, the upstairs should consider ABC first, and then (AB)C; At the same time, AC can be considered first, and then (AC)B; The results obtained should be the same. But:

ac, ac)b, the result is not the same as (ab)c

Here's what I thought:

abc, the forward probability is .

P(A forward, B forward, C forward) + P(A forward, B forward, C backward) + P(A forward, B backward, C forward) + P(A backward, B forward, C forward).

The backward probability is.

P (A backwards, B backwards, C backwards) + P (A backwards, B backwards, C forward) + P(A backwards, B forwards, C backwards) + P(A forwards, B backwards, C backwards).

Similarly, at ab(-c), the forward probability is .

P(A forward, B forward, C forward) + P(A forward, B forward, C backward) + P(A forward, B backward, C forward) + P(A backward, B forward, C forward).

The backward probability is.

P (A backwards, B backwards, C backwards) + P (A backwards, B backwards, C forward) + P(A backwards, B forwards, C backwards) + P(A forwards, B backwards, C backwards).

Answer: abc, the forward probability is; The backward probability is.

abc, the forward probability is . The backward probability is.

To put it simply:

a, b, c, the forward and backward probabilities of x are: ,.

a, b, c, the forward and backward probabilities of x are: ,.

Specific algorithms: forward, backward.

A , b , c, , ab, abc, so abc, so abc, x forward and backward probabilities are, in the same way, ab-c:

Forward, backward.

A , b , c, ab, ab-c, so ab-c, so ab-c, x forward and backward probabilities are, there are questions to ask.

Can this topic be done? Is there a mistake in writing?。。。

First of all, the probability that x satisfies each of the conditions does not tell us, and whether the conditions a, b, and c are independent of each other.

Let x be a random variable with a value of 0,1 (0 for forward, 1 for backward); a, b, and c are a, -a, and 0, respectivelyb,-b,0;c,-c,0 (0 means that the positive and negative conditions are not satisfied); What you are looking for is p, the conditional probability understands, and the backward probability is one minus the forward probability.

What is known is: p=; p=；p=

It should be possible to solve it using the conditional probability formula, but the conditions are not sufficient.

Conditional Probability Formula: p=p p

What do you think about the first floor and the second floor?,The formula of the column is your own thing.,I don't see any basis.。。。

That's right, the second floor is right.

60%, point p has the following possibilities: (-2, 4), (1, 1), (0, 0), (1, 1), (2, 4).

The parabola opens downward, the vertex coordinates are (1,6), past (0,5) points, you can roughly draw the image, when x=-2, y=-3, from the figure you can see that the first point is not inside, x=-1, y=2, the second point is inside. x=0, y=5, on the third point boundary, does not count. x=1, y=6, and the fourth dot is inside.

x=2, y=4, and the fifth point is also in it. So it's 3 5

The type of gift that two customers choose at random is 6*6=36 (each chooses its own, and there are 6 cases in total).

There happens to be the same thing: 4*3*2=24 (pick one out of the same giveaway, then pick two out of the remaining three, and assign two to two customers as giveaways).

k=24/36=2/3

It can also be reversed.

It just so happens that the opposite of one piece is that one is different + plus two are the same, and one is different, that is, 4 choose 2 to one of the customers 6 Both are the same, that is, 4 choose 2 to two customers 6

Thus k = 1-6 36-6 36 = 2 3

Put 3 identical balls into 3 different boxes at random, and the total release method = 3 + 3 * 2 + 1 = 10 (the sum of 3 balls represents the placement method of 3 balls in one box; Select 1 of the 3 balls and put them in a box, and put the remaining 2 balls into a box; One ball per box).

The maximum number of balls in the box is 1 The probability is the probability of one ball in each box: 1 10 balls are different in different boxes, and the total release method: 3 + 3 * 3 * 2 + 3 * 2 * 1 = 27 probability = 6 27

The difference between these two: 3 balls placed in the same box are 3 kinds, which are placed in different boxes. 2 boxes and 1 box, the first question is because the ball is the same, so only the box is selected; The second question is that the box and the ball are different, so you have to choose the ball.

1 ball per box, the first question ball is the same, so there is only 1 way to put it; The second question is because the box and the ball are different, the order of one fixed ball or box is fixed, and the other has 3 arrangements.

Are the balls the same? The box is still different, this one is the same as one of the previous two, right?

It is composed of 3 different numbers, 10 * 9 * 8 = 720 This question is similar to **, whether you are selected for the first time or the second time or the third time, the probability is 1 720, and the three times as long as there is one selection to win, so the probability of winning the lottery three times is 1 240 If you don't understand, then step by step.

First start k1=1 720

k=k1+k2+k3=1/240

For this problem, the slope of the straight line is equal to -1 and the straight line needs to meet the four cases of a=b, that is, =1, =3, =5, =7.

Any number taken from a set is a straight line, except in the case of a=b=0.

So, the probability k = 4 (5 2-1) = 4 24 = 1 6 b positive solution.

At least the probability of concentrating on the target: i.e. 1 - the probability of not hitting = 1-2 5*2 5=21 25=

A and B each shoot twice, and the probability that they hit the target twice in total: including A and B each hit once, A hit twice, B did not hit and A did not hit B twice = 117 400

Counting the points on MN, PQ, let's say A1, A2...., A6 is arranged from left to right. First, connect the points on A1 and PQ, and there is no intersection point.

The points on A2 and PQ are connected: the intersection point N1=6+5+4+3+2+1=21

The nodes on A3 and PQ are connected: the intersection point N2=2N1, you can get A4, A5, A6 and the points on PQ are connected: the number of intersections is: 3N1, 4N1, 5N1. (The number of intersection points resulting from the intersection of the following points and the 7 line segments of each previous point).

Then there are n=(1+2+3+4+5)*n1=15*21=315 (pcs).

But this question does not say that there is no common point at the intersection point. That is, the title does not exclude cases where things like a1b7, a2b6, and a3b5 intersect at a common point. So this multiple-choice question should be E instead

(1) 3 children may be 3 males, 3 females, 1 male and 2 female, or 2 males and 1 female. Because there is no birth order, the probability is 1 4

2) From the analysis of (1), it can be seen that the probability is 1 4

3) 3 males, 1 male 2 female, and 2 males and 1 female all meet the conditions of having at least 1 boy, so the probability is 3 4