Math probability problems, thank you for your help, thank you very much, thank you

It should be assumed here that the probability of each person passing the ball to each person is passed.

By default, each person has a 1 2 chance of passing to two other people, and each pass is independent of each other.

After n passes, the probabilities in the hands of these three are a n, b n, c n

and a 0 = 1, b 0 = 0, c 0 = 0

State transition: a = (b n + c n) 2

b_ = (c_n + a_n) / 2

c_ = (a_n + b_n) / 2

Considering that a n + b n + c n = 1 is constant.

And the similarity of the state transition equation can only be analyzed by taking a n, and the same can be obtained for other reasons.

a_ = (b_n + c_n) / 2

1 - a_n) / 2

Old. a_ -1/3 = -1 / 2 *(a_n - 1 / 3)

1/2)^ a_0 - 1 / 3)

a_n = (-1/2)^n * a_0 - 1 / 3) +1 / 3

Similarly. b_n = (-1/2)^ b_0 - 1 / 3) +1 / 3

c_n = (-1/2)^ c_0 - 1 / 3) +1 / 3

Bring in the initial value, get.

a_n = (-1/2)^n * 2 / 3) +1 / 3

b_n = c_n = (-1/2)^ 1 / 3) +1 / 3

At the same time, it is easy to find that when n tends to infinity, the three probabilities all tend to 1 3

Because each pass is a separate event, the probability of going to three people is equal. That is, after n passes, the probability of the ball going into the hands of all three people is equal, which is one in three. Hope it helps! Hope!

The probability of returning to the hands of the three is the same.

Second Pass: Find the probability of returning to Already and C.

Third Pass: Categorical Discussion.

The situation when the ball is returned to A: A passes B, B passes A; A to C, C to A. The probability is:

When the ball is returned to B's hand: A passes to C, C passes to B. The probability is:

When the ball returns to C's hand: A passes B, B passes C. The probability is:

It is found that the probabilities of the ball returning to A, B and C after n passes are p1 (n), p2 (n) and p3 (n) respectively

p1（n）=p2(n-1)*

p2(n)=p1(n-1)*

p3(n)=p1(n-1)*

Among them, p2(n)=p3(n), p1(n)+p2(n)+p3(n)=1 can only be counted step by step, I don't know how many n is.

If it is big, I think the probability of the ball coming back to A, B, and C should be 1 3.

According to probability, we can use conditional probability to calculate the probability of two gamblers winning at a point of (5,3), and further calculate the distribution of coins after winning.

Let gambler A and gambler match B be two gamblers, and their points are A and B, where A is the point of gambler A and B is the point of gambler B. In each round of betting, both gamblers' points have the potential to increase by 1 until one of them reaches 6 points, and he will receive the full amount of gold.

First, calculate the probability that Gambler A will win the early gear when the number of points is (5,3). Gambler A wins on the condition that he reaches 6 points first in subsequent games. Thus, we can list all the possible scenarios in which gambler A wins:

1.In the next round, Gambler A's value increases by 1 to 6. The probability of Gambler A winning at this point is 1 2.

2.In the next round, Gambler B increases by 1 to 4 points, and then Gambler A increases by 1 to 6 points. The probability of Gambler A winning at this point is 1 4.

So, at a point of (5,3), the total probability of Gambler A winning is 1 2 + 1 4 = 3 4.

Based on the probability of Gambler A winning, we can conclude that the probability of Gambler B winning is 1 - 3 4 = 1 4.

Therefore, the probability of Gambler A winning is 3 4 and the probability of Gambler B winning is 1 4. Gambler A gets 3 4 * 56 = 42 coins and gambler B gets 1 4 * 56 = 14 coins.

Summary. Can you send a text?

I don't know how to do a probability problem, so I ask netizens to help solve it, and write the result of this question clearly and send it over.

Can you send a text?

**The display is blurry.

Or you take a picture of the question.

Original question. The ** of this question has been sent to your mobile phone, please write the correct result of this question clearly, and take a clear ** and send it.

I have sent the original question** to your mobile phone, please help answer it.

Please help me write down the correct result of this question and send it to me.

It's not a complete question, you have to tell me what the A event is, what the first world is, so that I can know what it is interrelated.

You see.

Suppose the probability of being on hand A for the nth time is p(n).

If the ball is in A's hand for the nth time, then the ball must not be in A's hand on the n-1st time, but in someone else's hand, and will be passed to A on the next pass.

The probability of this person passing the ball to A is 1 4, therefore.

p(n) = ( 1 - p(n-1) )1/4 = 1/4 - p(n-1)/4 (n>1)

The value at n=1 is:

p(1) = 1/5

So p(n) = 1 4 - p(n-1) 4

1/4 - 1/4^2 + p(n-2)/4^2

1/4 - 1/4^2 + 1/4^3 -.1/4)^(n-1) +p(1)/4^(n-1)

1/4 - 1/4^2 + 1/4)^(n-1) -1/(5*4^(n-1))

1/4 * 1+1/4^(n-1)) / (1+1/4) -1/(5*4^(n-1))

1 + 1/4^(n-1))/5 - 1/(5 * 4^(n-1))

Mathematical induction proves that when 1) n = 1, the probability of the ball in the hands of 5 people is equal, all of them are 1 5, so p(1) = 1 5

2) Suppose that p(k)=1 5 when n=k, where k>=1.

If the ball is in A's hand at this time, the next round of passes will definitely not be in A's hand, and if the ball is not in A's hand at this time, the probability of the next round of ball holders passing to A is 1 4.

Thus p(k+1) = p(k) 0 + 1-p(k)) 1 4 = (1-1 5)*(1 4) = 1 5.

In summary, p(n) = 1 5 holds.

If it is 5 people, because it is a possible event, then the probability of the nth ball in each person's hand is 1 5 possibilities.

(1) First calculate the probability that the product of the 4 numbers on the 4 faces in contact with the desktop is not divisible by 4:

p'=(2 4) 4+4*1 4*(2 4) 3=3 16 (the first part is an odd product, the second part has a 2, and the others are odd).

Probability requested: p=1-p'=13/16

2) s can be taken: 0, 1, 2, 3, 4

When s=0, p(s=0)=(2 4) 4=1 16s=1, p(s=1)=4*(2 4)(2 4) 3=1 4s=2, p(s=2)=6*(2 4) 2(2 4) 2=3 8s=3, p(s=3)=4*(2 4) 3(2 4)=1 4s=4, p(s=4)=(2 4) 4=1 16 The distribution column is omitted.

Expectation = 0 * 1 16 + 1 * 1 4 + 2 * 3 8 + 3 * 1 4 + 4 * 1 16 = 2

As shown in the figure below, the number of white balls each time is either 0 or 1, and the probability of getting white balls for the second time is related to the first time you get white balls or black balls, so x,y are not independent:

If the vehicle stops, all the previous lights pass, i.e. the probability that all are 2 3. At this time, the light in front of you is a red light, which is a probability of 1 3.

When the car stops moving forward: 0 lights: 1 3

Over 1 lamp: (2 3) * (1 3) = 2 9

Over 2 lights: (2 3) 2*(1 3)=4 27 over 3 lights: (2 3) 3=8 27

If there are 30 days in a month, the probability of rain every day is 41 60.

1) (19 60) 10*(41 60) 20,The probability of rain in 10 days is multiplied by 20 days without rain!

2） (19/60)^1*(41/60)^29+(19/60)^2*(41/60)^28+(19/60)^3*(41/60)^27+..19 60) 10*(41 60) 20, "at most" means that there may be 1 day, 2 days, 3 days ......It rains for 8, 9, or 10 days, so add up the probabilities of all 10 scenarios!

3) For more than 10 days, you can use the formula 1 minus (2).

A month is counted as 30 days.

The probability of rain on a given day is.

The probability of not raining is.

1) It will rain for exactly 10 days, that is, multiply 10 times and multiply by 20 (2) As above, calculate the probability of rain on 0 days multiply 30.

The probability of rain on 1 day is multiplied by 1 and multiplied by 29.

The probability of rain on 2 days is multiplied by 2 and multiplied by 28.

The probability that it will rain for up to 10 days is the probability of rain for 0 to 10 days and the probability of (3) for more than 10 days is the probability of 1 - for up to 10 days (the result of 2).

A month is counted as 30 days.

The probability that it will rain every day is a=, and the probability of not raining is b=

1 Probability of rain for exactly 10 days = c(30,10)*a 10*b 202 Probability of rain for up to 10 days including 0 days, 1 day, 10 days of rain = b 30+c(30,1)*a*b 29+·· c（30，9）*a 9*b 21+c（30，10）*a 10*b 20

3 The probability of rain for more than 10 days is to subtract the probability of the second question from 1, because the probability of rain for up to 10 days = 1 - the probability of rain for more than 10 days, these two events are opposite events.

This problem is calculated according to the binomial distribution, that is, the probability of rain every day is equal, the method is not complicated, but the calculation is more complicated).

1.Probability of rain in a day:; Probability of no rain: (;

2.It rained exactly 10 days :(;

3.Up to 10 days of rain: exactly 1 day to exactly 10 days;

4.More than 10 days of rain: exactly 11 days to exactly 3