Organize recent notes.
It will be updated from time to time. If you have copyright issues, please contact the author and relevant content will be deleted immediately.
Introduce
\ (eg1 \)
Four players of A, B, C and D do the pass practice. The ball is first passed by A, and each person passes the ball to one of the other three people with equal probability. Let \ (P_n \) indicate the ball after \ (n \) passes a return to the hands of probability, seeking:
(1) \ (P_2 \) . the value
(2) \ (P_n \) (to \ (n \) expressed \ (n \) after the passing of time the ball landed armor In hand.
Answer:
In a pass, the ball can be passed from [Not A] to [A], then [Probability of the current A] = [Probability of the last time the ball was in B and passed to A] + [Last Probability of the ball in C and passing to A] + [Probability of last ball in D and passing to A]:
and so
Next, add the coefficient:
The first pass must be in the hands of A, so \ (P_1 = 0 \)
so
\ (eg2 \)
The three players A, B, and C take turns throwing a cube with a uniform texture. The rules are as follows: If someone rolls \ (1 \) points one time, the next time they continue to roll, if they roll other points, then The other two people caught Yan and decided who would throw it, and the first one was thrown by A. Assume that the probability of throwing from the first \ (n \) is \ (p_n \) , and the probability of throwing by B or C is \ ( q_n \) . Find the general formula of the sequence $ {p_n} $.
Answer:
Consider: [Probability of being A this time] = [Last time is 1 o'clock] + [Last time is B not in 1 o'clock, you have captured the armor] + [Last time you are not in 1 o'clock, you have captured the armor bingo】
Simplify:
We know that during the \ (n-1 \) game, the sum of the probabilities of the three players throwing dice is \ (1 \) , so:
So as long as the formula is connected, you can get
Solutions have to
Advanced (However, it is still very simple)
\ (eg3 \)
Some people play the game of throwing tetrahedral dice and walking checkers. It is known that the four sides of the tetrahedral dice are printed with \ (A, B, C, D \) on the board marked with the station \ (0 \), station \ ( Station 1 \), station \ (2 \) ,..., Station \ (100 \) . A piece starts at station \ (0 \) . Each time the player rolls the dice, if the dice is In the \ (A \) plane, the pawn jumps forward \ (2 \) , if the dice is a side in \ (B, C, D \) , the pawn jumps forward \ (1 \) , until When a piece jumps to station \ (99 \) (Victory Base Camp) or station \ (100 \) (failure Base Camp), the game ends. Let the probability that the piece jump to station \ (n \) be \ (P_n \ ) \ ((n \ in \ mathbb (N)) \) .
(1) Find the relationship between \ (n \) and \ (P_n \) .
(2) Find the probability of winning and losing the game.
Answer: The
current location \ (n \) can be reached from \ (n-1 \) or \ (n-2 \) , so
The characteristic equation is:
Solutions have \ (x_1 = 1, x_2 =
- \ frac {1} {4} \) undetermined coefficients:
Bring in two special values \ (0 \) and \ (1 \) into the solution \ (A, B \)
Solved \ (A = \ frac {4} {5}, B = \ frac {1} {5} \)
So
Then the probability of winning is:
The probability of failure is:
In short, do not bring back the general formula of \ (P_n \) to calculate when it fails , because you have won when you are in the \ (99 \) grid, and will not continue to go back, which does not meet \ (P_n \ ) The prerequisite for the establishment of the general term \ (n \ leq 99 \) , if he directly asks you the probability of failure and then you split without noticing you
\ (eg4 \)
Now a herd of cattle on whether consider themselves handsome random survey. If they are not considered handsome remember \ (1 \) points, if followed consider themselves handsome remember \ (2 \) points per cow think you are handsome probability \ (\ frac {2} {3} \) , and the event that each cow thinks he is handsome is independent of each other. During the random questionnaire survey of all cows, the cumulative score that has been investigated is exactly \ (n \ ) The probability of the score is \ (P_n \) , find the general formula of the sequence $ {P_n} $.
answer:
Method 1:
[Probability of getting current points] = [Probability of getting \ (n-1 \) points and then \ (1 \) points] + [Getting \ (n-2 \) points and then \ (2 \) points The probability】
Or we can think in another direction;
Method 2:
Because there is only one situation when you do n’t get \ (n \) points, that is, when you interview n-1 points, the next cow you think is handsome, skip \ (n \) at this stage. Too. That is:
1- [Probability of getting current score] = [Probability of getting 2 more points when getting n-2 points]
which is:
The two recursive methods lead to the same path and get the solution:
\ (eg5 \)
There are \ (8 \) balls in the bag , including \ (5 \) white balls and \ (3 \) red balls. These balls are exactly the same except for the color. One ball is randomly taken from the bag. If the red ball is taken out , Put it back in the bag; if the white ball is taken out, the white ball will not be put back, and another red ball will be put into the bag, after repeating the above process \ (n \) times, the number of red balls in the bag Record it as \ (X_n \) . Find the mathematical expectation of the random variable Xo \ (E （X_1 \) ) about the expression of \ (n \) .
answer:
It is easy to know that $ X_n $ represents the number of red balls. During the process of removing and putting them back, the red balls may only increase but not decrease. When the bag is full of red balls, then take it out and put it back to play a bag of \ (8 \) red balls. So the possible value of \ ( X_n \) is \ (3,4,5,6,7 , 8 \) .
If there are currently \ (k (k \ in [4,8]) \) red balls, then it can be transferred from the last \ (k \) red ball or \ (k-1 \) red ball and Come. Then
In particular
After all, there are only \ (3 \) red balls at the beginning, it is impossible to transfer from the state of \ (2 \) red balls.
We can write according to the desired definition:
According to the sum of the probabilities of all independent events \ (1 \)
It can be guessed that \ (E (X_n) \) can be obtained by \ (E (X_ {n-1}) \) linear "patchwork", so the coefficient to be determined, set
Substitute into the above solution and get \ (x = \ frac {7} {8}, y = 1 \)
then
Since \ (E (X_0) = 3 \)
then
Of course, we can also change the way of thinking and use the expectation of the difference value to find the original expectation.
We know that the possible value of \ ( X_ {n + 1} -X_n \) is \ (0,1 \) .
Then expect:
There is still a random variable, and another layer of \ (E \) .
Set several layers of \ (E \) are the same, whether it is \ (E (E (X ()) \) or \ (E (E (E (E (E (X)))) \) is equal to \ (E (X) \) .
and so
Get recursive
The result is still the same.
egEX Some doctors collaborated with some physicists and recently discovered a pair of microorganisms that reproduce in a special way. The male microorganism is called double phage \ ((diphage) \) and has two receivers on its surface; and the female microorganism is called triphage \ ((triphage) \) and it has three receivers:
When the cultured tissues of double phage and triple bacteriophage are irradiated with \ (\ psi \) particles, there is exactly one receiver on the bacterium to absorb the particle, and each receiver is equally possible . If this is a strain of double bacteriophage The receiver, the double bacteriophage is transformed into a three bacteriophage; if this is a receiver of the three bacteriophage, the three bacteriophage will split into two double bacteriophage, so that if an experiment starts with one double bacteriophage, the first A \ (\ psi \) particle turns it into a three-phage strain, the second particle turns this three-phage strain into two biphage strains, and the third few particles turn the two One becomes a three bacteriophage, the fourth \ (\ psi \) particle either hits a forest of double magnets, or hits a triple magnet, so that two strains of three bacteriophages are obtained (probability \ (p_1 \) ), Or get three double bacteriophages (probability \ (p_2 \) ). Now start from a single double bacteriophage and irradiate its cultured tissue with a single \ (\ psi \) particle \ (n \) times,
\ ((I) \) Find the value of \ (p_1, p_2 \) ;
\ ((II) \) Solve the following problems:
\ ((i) \) After \ (n (n \ in \ mathbb {N ^ *}) \) times of irradiation, there are \ (n + 2 \) identical receivers. Set a random variable \ (X_n \ ) Represents the number of double bacteriophages present, and the random variable \ (Y_n \) is the increase in the total number of receivers (may be negative) when the \ (n + 1 \) th \ (\ psi \) particle is irradiated. the \ (Y_n \) of the distribution column, and determining the \ (Y_n \) mathematical expectation \ (EY_n \) (results containing \ (EX_n \) , \ (EX_n \) is a random variable \ (X_n \) mathematical expectation ).
\ ((ii) \) finds the linear relationship of \ (EX_n (n> 4) \) about \ (n \) .
(Answer to be continued)