Who will help me solve the problem about probability in Mathematics Compulsory 3?

Well, it's simple, it's OK

Question 1: The probability of drawing three out of it is c6 3 (I can't play it, you know what it means......One of them should be C3 1 and the other two are drawn in the second and third class, so it is C3 2

The probability is: c3 1*c3 2 divided by c6 3=9 20 Second question: The total probability of drawing three out of them is c6 3

Two first-class products are drawn from three first-class products, with a probability of c3 2 and the other one is drawn from the second and third classes, with a probability of c3 1

So it is: c3 2 * c3 1 divided by c6 3 = 9 20 Third question: The total probability of drawing three of them is c6 3, there is no third class, so all three are in the first and second class The total number of first and second class is 5, so three of them are drawn from it, and the probability is c5 3

Probability: c5 3 divided by c6 3 = 1 2

1) There are (3c1)*(3c2)=9 kinds of selections, so the probability is 9 (6c3)=9 20

2) The total selection is (3c2)*(3c1)=9, so the probability is 9 (6c3)=9 20

3) The total selection is 5c3=10, so the probability is 10 (6c3)=1 2

Solution: According to the meaning of the question, get.

1)p1=c31*c32/c63

2)p2=c32*c31/c63

3)p3=1-(c31*c21+c32*c20+c30*c22)/c63

Answer: (1) The probability of having a first-class product is 9 20;

2) the probability of having exactly two first-class products is 9 20;

3) the probability of not having a third-class product is 1 2;

Solution: (1), first draw a circle can be painted red, blue, then draw a triangle, and then draw a rectangle are also two results, all the results are equivalent to a coin tossed three times in a row There are 8 basic situations, for.

Round red three blue long blue round blue three red long blue round blue three blue long red round blue three blue long red round blue three blue long red round blue three red long red round red long red round red three red long red round red long red this tree diagram you go to draw yourself, this is not easy to draw.

2）p(1)=1/8

p(2)=4/8=1/2

p(3)=4/8=1/2

p(4)=6/8=3/4

red-blue-red, red-blue-blue, red-red-blue, red-red-red, blue-blue-blue, blue-blue-red, blue-red, blue-red, blue-red.

At least one of them has a chance of being elected.

1 minus the probability of not having one.

None of the girls have a chance of being to:

c(2,7)/c(2,10)=7/15

So the probability of at least one being elected is 1-7 15 = 8 15c(m,n) is the number of combinations of n out of n.

8/15;First calculate the probability that all boys will go, that is, (2 out of 7) (2 out of 10) to get 7 15, and then subtract 1 to get the answer. It's just a common way to find "at least" questions.

Answer D, Xiaohua can only meet when he arrives between 1:30 and 1:50, accounting for the total time

1 red round, red three, red long |Red round, blue three, red long |Red round, blue three, blue long|Blue round, red three, red long |Blue round, blue three, red long |Blue round, blue three, blue long|Red round, red three, blue long|Blue round, red three, blue long.

2 1 1 1 1 8 in 1.

2 2 1/2.

2 3 1/2.

2 4 three-quarters.

1.There is exactly 1 first-class product: c(1,3)*c(2,3) c(3,6)=3*3 20=9 20

2.There are exactly 2 first-class products: c(2,3)*c(1,3) c(3,6)=3*3 20=9 20

3.There is no first class: c(3,3) c(3,6)=1 20

What does the question mean? I don't understand. Please be clear.

As above, the four positions are numbered.

1, a on the edge means a in or position.

2, A and B are both on the edge of the denote and can only row A and B3, A or B on the edge means A is on the edge, B may or may not be on the edge; or b on the edge, a may or may not be on the edge.

4, A and B are not on the side, A and B can only be in the and position I have never used opposing events, as long as you understand the meaning of the topic. Confrontational events can get answers or trouble.

The first question a has c(2,1)*a(3,3)=12 on the side, and the second question a and b are on the side.

a（2,2）*a（2,2）=4

The third question A or B is divided into three cases on the side.

1) A is on the edge B is not on the edge: C(2, 1) * C(2, 1) * A(2, 2) = 82) B is on the edge A is not on the side: C(2, 1) * C(2, 1) * A(2, 2) = 83) ab are all on the edge:

a(2,2)*a(2,2)=4 is added in three cases: 8+8+4=20

Neither question A nor question B is on the edge of the fourth question: a(2,2)*a(2,2)=4 So it seems that the second and third questions are not opposite events, because there is a common image (A and B are on the side), and the fourth and third questions do not intersect, so the fourth and third questions are opposite events. There may be an error in the answer.

If A is true, then B is also true, so they have an intersection; But if C is true, D must not be true, and vice versa, so they do not intersect, so they are opposed.

1. Set a to mean that it can't be opened the first time, and b means that it can be opened the second time.

If the door cannot be opened and thrown away, the probability required is p(ab)=p(a)p(b|).a)=2/4*2/3=1/3

If the tried key is not thrown away, the required probability is p(ab)=p(a)p(b|).a)=2/4*2/4=1/4

Ask the probability of opening the second time, so the first time it is not opened, i.e. c(1,2) c(1,4).

If you can't open the door, throw it away, that is, take the key that can open the door from the 3 keys for the second time, i.e. c(1,2) c(1,3).

So the result is c(1,2) c(1,4) multiplied by c(1,2) c(1,3)=1 3

The second case is that the key is not thrown away, that is, the second time the key that can open the door is taken from the 4 keys, i.e. c(1,2) c(1,4).

So the result is c(1,2) c(1,4) multiplied by c(1,2) c(1,4)=1 4

If event a means that the first time it cannot be opened, and event b means that it can be opened the second time and the door cannot be opened, then the required probability is p(ab)=p(a)p(b|).a)=2/4*2/3=1/3

If the tried key is not thrown away, the required probability is p(ab)=p(a)p(b|).a)=2/4*2/4=1/4

University knowledge.

Solution: From the meaning of the question, this question is an equal possible event, the event contained in the test is that two people each have 6 different methods, a total of 36 results, the event that meets the conditions can be from each floor, a total of 6 results, the probability of two people leaving the elevator on the same floor is 6 36 = 1 6, then the probability of 2 people leaving on different floors is: 1-1 6 = 5 6;

So the answer is: 5 6

Good luck with me.

Method 1 (positive and difficult) :p=1-1 6*1 6*6=5 6 Method 2: p=1 6*5*6*1 6=5 6 Method 3: Draw a treemap, there are 6 kinds that meet the requirements of the question, and there are 36 kinds in total, so p=6 36=1 6

When it comes to exams, method 1 is the most preferable, method 2 is more demanding in thinking, and method 3 is more troublesome but the easiest.

When you can't do it in the college entrance examination, method three is the best way!

Good luck!

Let's first discuss the probability of two people starting from the second layer and going all the way to the seventh layer, and leaving the same layer together p=1 6 *1 6 6*6 6

So the probability p'=1-p=5 6 is sought

There are 1 cases where 6 of the 10 numbers are exactly the six numbers that win, and then suppose that 5 of the 6 numbers are selected, and there is no 6, and 1 of the remaining 4 numbers is selected, then there are 4 such situations;

In the same way, suppose there are no 5, no 2, no 3 in 6 numbers... There are 4 such situations.

So there are 25 ways to win the lottery, 6x4+1

Choose 6 out of 10 numbers, there are 210 according to the combination formula, the probability is 25 210 = 5 42, choose d

Let a mean that B means that the second can play.

In the case of the energy gate being thrown away, the probability p(ab)=p(a)p(b|) is requireda)=2/4*2/3=1/3

In the case of test key throwaway, the probability p(ab)=p(a)p(b|) is requireda)=2/4*2/4=1/4

Beijing and the state and the attack are high.