High score, probability theory solving, urgent, thank you

5. 1) by. The formula for uniform distribution is f(x)=1 (b-a) f(x)=x-a (b-a).

It can be seen that f(x)=1 3 ,02)p(max(x,y) 1) is equivalent to p(x 1, y 1 ) and because x,y is independent.

So p(max(x,y) 1) = p(x 1)*p(y 1) = 1 9 six 1) p(x,)=p(x,1)+p(x,2)+p(x,3) x=1,2

p(.,y)=p(1,y)+p(2,y) y=1,2,3 due to p(1,.)p(.,1)=,1)

So XY is not independent.

2)cov(x,y)=e(xy)-e(x)e(y)=d(x)=

d(y)=x,y)=cov(x,y) d(x) 1 2d(y) 1 2=ls you don't subtract the expectation when calculating the variance.

Formula: Can be rigorously proven.

If x obeys ex=u, dx=δ 2

For: x pull = 1 n xi

e(x pull) = u, d(x pull) = δ2 n

So, for this question, n=10, with selected b

If you choose b, you will expect e(ax)=aex, and if you have variance, d(ax)=a2 *x, so the expectation is still the expectation of n, but the variance should become 1 10.

According to the distribution function.

The solution process is as follows.

If you don't understand, please ask, satisfied.

f(x,y)=x+y (0<=x<=1,0<=y<=1)

Probability density. f（z)=p(z<=z)=p(max<=z）=p(x<=z,y<=z)

When z<=0, f(z)=0

When 0 is = 1, f(z) = 1

So f(z)=3z 2, probability density.

f（z)=p(z<=z)=p(min<=z）=1-p(min>z)=1-p(x>z,y>z)

When z<=0, f(z)=0

When 0z,y>z)=1- [z,1]dx [z,1](x+y)dy=1- [z,1](x+1 2-xz-z 2 2)dx=1-(1-z-z 2+z 3)=z+z 2-z 3.

When z>=1, f(z)=1, so f(z)=1+2z-3z2,0

13 questions, e(x)=pi)(xi) =2) *e(4x +6)=4e(x)+6=12.

In question 14, the edge distribution of y is, p(y=-1)=, p(y=0)=, p(y=2)=0+.

The edge distribution of x is, p(x=0)=, p(x=1)=, p(x=2)=.

FYI.

c can use Wayne ** Duel.

c Bernoulli distribution, c(n,0)p 0*(1-p) nd is the probability of 1 2

b Same as question 3.

Before drawing the registration form for each region, a region should be randomly drawn, and the probability of one, two, and three regions is 1 3

Your x should be 1 9, y 1 6, and z 5 72

The rest is right.

The sample space of the above solution is chaotic, and the sample space is (ai, bi, ci) ai (i region), b1, b2 (male and female are drawn for the first time), and c1 and c2 (male and female are drawn for the second time).

What you are asking for above is p(x)=p(b2|a1c1)p(m)=p(a1)p(b2|a1c1)+p(a2)p(b2|a2c1)+p(a3)p(b2|a3c1)

m≠x+y+z

Question: It's not a question, is it [covering face].

Ask the bottom multiple-choice question I asked.

Probability theory, I think the main thing is to test a logical thinking ability, you have to figure out what event is talked about in this question before solving it.

First of all; The probability of an event in each region is the probability of the first draw of a girl multiplied by the probability of the second draw of a boy.

For example, if the first region is 3 10 times 7 9, you don't need to find p(b|a), because it is not played back after being extracted.........And then; Girls were drawn first and then boys were drawn in all three regions, which should be the average of the probability of the event in the three regions.

This can be understood in this way, the probability of using region one to do spot checks is 1 3, and the probability of sampling women and then men in region one is 7 30, the same for regions.

The same is true for the second and third parties.

The above solution to this problem can be understood as follows: 1) Take p(b1|) in a regiona1)=p(a1b1)÷p(a1)=1/3

2) Take p(b2|) in region 2a2)=p(a2b2)÷p(a2)=1/2

3) Take p(b2|) in the three regionsa2)=p(a2b2)÷p(a2)=5/24

Your m=x y z is: "If a=a1 a2 a3, b=b1 b2 b3, then p(b|."a)= p（b1|a1)+ p（b2|a2)+ p（b3|a3)"

Obvious error.

What needs to be explained in the question is x1,..XN is an independent sample from the population.

First, calculate the variance of the x-means baixbar.

var xbar = 1/n^du2 * 1+1+1+..1) = 1/n

The correlation ZHI coefficient of xbar and arbitrary xi is then calculated, which is additive by the DAO.

cov (xbar ,xi) = cov(x1/n, xi)+ cov(x2/n, xi) +cov(xn/n, xi)

0+0+..0+cov(xi/n, xi)+0+0+..0=1/n * varxi = 1/n

So continuing by additivity, cov (xi-xbar, xj-xbar) = cov (xi, xj) -cov(xi, xbar) -cov(xbar, xj) + cov(xbar, xbar) = 1 n-1 n-1 n = -1 n

var(xi-xbar) = var xi+varxbar - 2cov(xi,xbar) = 1+1/n - 2/n = (n-1)/n

The correlation coefficient is.

RHO = covariance of both Multiplied by square = -1 n (sqrt(n-1 n) *sqrt (n-1 n)) = -1 n *n (n-1) = -1 (n-1).