Math Probability Thanks! 10

Because there are only 4 options for each question, the odds of getting it right are the same, and for 20 multiple-choice questions, 5 questions will be answered correctly, and because 60 of them are correct to pass the exam, so x60 = 15. I don't know if it's right.

This is a probability question, and the pass can be on 12 to 20 questions. Take 12 questions as an example, which is equivalent to 20 questions and then choose 8 questions to make mistakes, that is, c (20 subscripts and 8 superscripts). This is calculated as 1*2*3*4....*8/(8*9*10*…*20)。

Calculate multiplying in (1, 4), 12, (3, 4), 8. The latter ones just change the corresponding numbers according to the rules, and then add them up. Upstairs calculates that there are only 12 questions, and then the probability that all 12 questions are correct.

This is the probability of high school.

That is to say, you can answer 12 questions correctly, the probability of answering each question correctly is 1 4, and the probability of passing is 1 4 to the 12th power.

This one is very simple: 20 questions as long as 12 of them are answered correctly.

This is a typical permutation and combination question: choose 12 out of 20 questions and answer correctly 12 20c*1 4 1 4 * 1 4 *1 4 *....12 pcs.).

It worked, wow. Others have questions about their answers, and they only calculate the probability of getting the first 12 questions right, but don't take into account the fact that the last 12 questions are correct.

There are 4 options, which means that he has a 1 in 4 chance. Less than 60%. The probability of passing the exam is 0

Over or not. Believe it or not, they're all right, and that's at least 12 ways, plus the probability of the tract, independent repeat experiments, and the algebra formula is.

Khanna, this kind of question, are you going to prepare for yourself?

What is n? Calculated with burial years n=6.

All combinations have a total of c(10,2) =45 species, the number of species that get the same juice is c(6,2)+c(4, liquid ridge 2)=15+6=21, probability 21 45 = 7 15, the number of species to get different juices is c(6,1)*c(4,1) =24 (or 45-21 = 24), probability 24 45 = 8 15.

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:

This first question is not much different from the sets we learned in middle school, which means that when we understand the operation of events, we can understand and remember according to the operation of sets. Of course, events also have their own unique properties. Therefore, this first problem is basically to use the operation of the set to write the result, and there may be a part of the event that is deformed first, and then the result is written.

The second question should examine the arithmetic laws of events, including commutative, distributive, associative, and Morgan's laws.

10 cases: 4 + 6, 5 + 5, 6 + 4 a total of 3 cases.

And the total situation: 6 * 6 = 36 (species).

Probability p=3 36=1 12

The first one is at least 4, so the probability that the first one gets greater than 4 is 3 6 = 1 2

For the number of the first dice, obviously, the second dice have only one number, and the corresponding probability is 1 6

So it's 1 2*1 6=1 12

There are two scenarios. 1. The body is gone.

Row the last system, 2 bodies, the last row. When the body is at the end, there is C31 (choose one of the three) * A33 (three rows left) The result is recorded as P1 When the body is not at the end. There is C31 (in addition to the first and last three to choose a row) * C21 (the technique is not adjacent to the body.

So there are only two positions) *A33 (any row of the remaining three subjects) The result is denoted as P2 and there are a total of A55 arrangements. Write it as p3...So (p1+p2) p3=9 20

Total a5 5 = 120

The technical class ranks first 3*a3 3=18

The technical class is not ranked first, a3 3*a3 2=36

p=(18+36)/120=9/20

The denominator is 5*4*3*2*1, and the numerator is (4*2+1)*a33The result was 9 20

The full permutation is a5 5 = 120

When E is in first place, A4 = 24

When e is not in the first place and d, e is adjacent to c3 1*a3 3+a3 3=24, then the probability is 120-24-24 120=3 5

Since x,y obeys a two-dimensional normal distribution (1,0,32,42,,, it can be inferred from the textbook that x obeys a normal distribution (1,32) and y obeys a normal distribution (0,42), and the correlation coefficient between x and y is .

1) z = x 3 + y 2, so it is easy to know from the additivity of the mean and variance of the normal distribution (textbook theorem) that z obeys the normal distribution (1 + 0, 32 + 42), i.e. (1, 74).

2) xz refers to the correlation coefficient between the random variable x and z!! Let xz=a, then a=cov(x,z) [(d(x)) square) *d(y))]], d[x] and d[y] are 32 and 42, so just find cov(x,z). And cov(x,z)=cov(x,x3+y 2)=cov(x,x 3)+cov(x,y 2)=(1 3)*d(x)+(1 2)*cov(x,y), and since the correlation coefficient between x and y is, the set of formulas can directly find cov(x,y), and d(x) is known, so cov(x,z) can be calculated.

3) According to the result of the second question, if the correlation coefficient xz of x,z is equal to 0, then x,z are not correlated; Not equal to 0 is relevant. The correlation and independence of normally distributed random variables are equivalent, so it is possible to determine whether x and z are independent.

To make flexible use of the nature of these random variables, this kind of question will generally not give you points, otherwise is it still an exam question? Hehe.

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.