Probability questions, speed helps! How to do this probability problem? Please help, thank you!

1. Everyone has the possibility of hitting all three shots, as long as one person does not hit all three shots, both of them have a chance to shoot.

If A starts first and the probability of hitting all three shots: 1 2*1 2*1 2=1 8, the probability of not hitting all of them is 7 8;

If B starts first and the probability of hitting all three shots: 3 4*3 4*3 4=27 64, the probability of not hitting all of them is 37 64

They decide who shoots first through coins, and the probability of A or B shooting first is 1 2, so the probability that both have a chance to shoot: 1 2*7 8+1 2*37 64=93 128

2, exactly 4 grab the end, if A starts first, there may be.

A, A misses first, B then hits all 3 shots, probability: 1 2*3 4*3 4*3 4=27 128

B, A first shot 1 shot, second shot missed, B then hit 2 shots, probability: 1 2*1 2*3 4*3 4=9 64

C, A first shot 2 shots, the third shot misses, B then hits 1 shot, probability: 1 2*1 2*1 2*3 4=3 32

If B starts first, there may be.

A, B misses first, A then hits all 3 shots, probability: 1 4 * 1 2 * 1 2 * 1 2 = 1 32

B, B first shot 1 shot, the second shot misses, A then hits 2 shots in a row, probability: 3 4 * 1 4 * 1 2 * 1 2 = 3 64

C, B first shot 2 shots, the third shot misses, A then hits 1 shot, probability: 3 4 * 3 4 * 1 4 * 1 2 = 9 128

So the probability of four shots ending exactly is: 1 2 (27 128 + 9 64 + 3 32) + 1 2 (1 32 + 3 64 + 9 128) = 19 64

1. The order in which both people have a chance to shoot can be: AB, A, B,

The various probabilities are:

A and B: 1 2*1 2*1 2. The first 1 2 is the probability that a coin is tossed and shot by A, the second 1 2 is that A misses, and the third 1 2 is the probability that B tosses a coin and B shoots.

A, A, B: 1 2*1 2*1 2*1 2*1 2*1 2A, A, A, B: 1 2*1 2*1 2*1 2*1 2*1 2*1 2*1 2*1 2*1 2*1 2

B B A: 1 2*1 4*1 2*1 4*1 2 B B B A: 1 2*1 4*1 2*1 4*1 2*1 4*1 2 2 My method looks a bit complicated, and I haven't figured out the method of bento yet...

It's been too long since I got the probability.

2, just the probability of the end of the four shots, wait for me to be free and then sort out my thoughts.

e e e I read the topic wrong. Tears run...

The probability that the target will be hit twice is equal to.

p (A, B, C) + P (A, B, C) + P (A, B, C) Here, "pull" indicates its opposite.

Obviously, their shots are independent, so, p(a, b, c) = p(a) p(b) p(c).

p(A, B, C) = p(A) p(B) p(C)p(A, B) p(A, B, C) = p(A) p(B) p(C) The number of incoming is worth: p=

Distributed independently like the one above, that's what I did.

The probability of taking the Black Sox for the first time is 3 5

The probability of taking the black socks for the second time (3*2-1) (5*2-1) = 5 9 to take out a pair.

Black socks probability 3 5 * 5 9 = 1 3

The probability of taking white socks for the first time is 2 5

The probability of taking white socks for the second time (2*2-1) (5*2-1)=1 3 Take out a pair of white socks.

Sock probability 2 5 * 1 3 = 2 15

Chance of taking out a pair (2 of the same color is only for a pair) 1 3 + 2 15 = 7 15 Chance of red top and white trousers 1 3 * 1 2 = 1 6 Then I will add.

(black, black) 6 (white and white) 12 (black and white) 12 (white and black) 15 + 6 = 21

2.Red-blue, red-white, yellow-blue, yellow-white, blue-blue, blue-white.

1/6ok?I was about to take the high school entrance exam in my third year of junior high school.

（1）..There are a total of four situations that arise in this question. It's all invalid and valid, b invalid invalid, b valid and both valid.

Question (1) At least one valid contains three cases. So just subtract 1 from 1 that won't happen.

The probability of both invalid is (

The probability that at least one of the two alarm systems is valid.

2)..The probability of b failure is a probability of being effective is.

Therefore the probability that b is out of order and a is valid x =

The conditional probability is calculated as follows:

First question. r is 10, and the two black balls are removed on the first and third times, p=1 c24=1 6

Second question. e=2^1+2^2+..2^(n+1)-(2^1+2^2+..2^(n+1))/n+1)=(2^(n+2)-2)n/(n+1)

There is a problem with this question, the maximum score is 4 points in total, how can you find the probability of 10 points? That's 0.

I feel like you've copied the wrong question, and according to your question (1)n=m=2, then r must =4 p(r=10) is 0, and I guess it may be the ith time you get black and get 2i points, then p(r=10)= 1 3 because only the first and fourth are black balls, or the second and third are black balls, and you can score 10 points.

(2) Ask me not to wait for your feedback to see if the question was copied incorrectly.

On the reverse side, the probability that the national emblem is facing up once is (to the fifth power.

Reduction is 1- (to the fifth power.)

Is it okay to ask a question?

This is 101, not 100.

Misread, it's 100

Question 1: The probability of waiting for a lottery to get super excellent is: 40%;

The probability of getting the best in the second draw is: 39%;

The probability of winning the super-excellent twice at the same time is: 40%x39%=, then the probability of not drawing the super-excellent at the same time is 1-156 1000=

The probability of waiting for the two best to be drawn at one time is: 35%;

The probability of getting the best in the second draw is: 34%;

The probability of winning the super-excellent twice at the same time is: 35%x34%=, then the probability of not drawing the super-excellent at the same time is the probability of waiting for the quasi-two excellent draws at the same time: 25%;

The probability of the second draw is 24%;

The probability of winning the quasi-two advantages at the same time is: 25%x24%=6%, and the probability of not drawing the quasi-two advantages at the same time is 1-6%=94%.

Correct the "two excellent" points:

The probability of waiting for the two best to be drawn at one time is: 35%;

The probability of winning the two excellent results in the second draw is: 34%;

The probability of drawing two excellent at the same time is: 35%x34%=, then the probability of not drawing two excellent at the same time is to ask questions, the teacher helps me draw a tree diagram, I just learned about probability, a little dizzy to answer sorry, classmate, there is no pen and paper around now, wait for a while and then draw a second question for you: choose super excellent 1000

(1) Probability of both pieces being bad = 20%*10%=

The first 2 of the 3 people get 2 bad, and the last person is not, the probability = the position is interchangeable, then the answer =

2) What is the distribution column? I guess it's a picture.,Draw it yourself.,Maybe it inspires you.,I don't understand.,But I can't help the answer anyway.。

2. Good probability = 80% * 90% =

2 Bad = 1 Good 1 Bad =

Math expectation x = pcs.

I'm sorry that the answer is missing a distribution column.,I see your speed to solve.。。。 50 min!。

Signature file: I sincerely hope that communism will come as soon as possible! This is just the first step for humanity!

1. P (exactly two people get two unqualified).

c3,2 * c2,1)^2 * 1-90%)^2+(1-90%)*1-80%)+1-80%)^2] *c2,1 * 90%+c2,1 * 80%]

2、p(x=0)=90% *80% =p(x=1) = c2,1 * 1-90%) 80%+ c2,1 * 90% *1-80%) =

p(x=2) = (1-90%) 1-80%) =e(x) = 0 * 1* 2* =

The first question: The probability of shooting down an aircraft with a single shot is:

The probability of shooting down the aircraft with the second shot is: (

The probability of shooting down the aircraft with the third shot is: (

The probability of being shot down after three shots is:

Second question: The plane was shot down and hit by two bombs.

There are two scenarios:

1: The first bomb hit the plane and did not shoot down, the second bomb shot down, this probability is: (2: one of the first two bombs hit but did not shoot down the plane, and the third bomb shot down the plane, in this case there are also two situations:

The first bomb hits the plane and does not fall, the second bomb misses, and the third bomb shoots down, the probability is: (state band: the first bomb misses, the second bomb hits the plane and does not fall, and the third bomb shoots down, the probability is:

The total probability is the sum of these three cases: the method is definitely like this, and the calculation may be biased) to ask if you don't understand.