Find the probability problem of mathematical expectation. Probability theory, finding mathematical e

The probability of each die appearing 1-6 is 1 6

The probability of a product of 1-6 squares is 1 36

The probability of other outcomes is 1 18, because for example, 1 and 2, there are two possibilities for this situation, dice 1 is 1 point or dice 2 is 1 point.

e is then equal to the product of each possible multiplication, multiplied by their respective probabilities.

This is a negative binomial distribution, let x be the number of times you need to have 3 heads, y be the number of failures, r=3

There is a formula, ey=r(1-p) p=3

ex=ey+r=6

My drop probability theory is also super bad, refer to it.

If the product of 1 is x, then x=1,2,3,4,5,6,8,9,10,12,15,18,16,20,24,25,30,36,p(x=1)=1 36 p(x=2)=1 18

p(x=3)=1/18 p(x=4)=1/12 p(x=5)=1/18 p(x=6)=1/9 p(x=8)=1/18 p(x=9)=1/36 p(x=10)=1/18

p(x=12)=1/9 p(x=15)=1/18 p(x=18)=1/18 p(x=20)=1/18 p(x=24)=1/18 p(x=16)= 1/36

p(x=25)=1/36 p(x=30)=1/18 p(x=36)=1/36

The expectation is =1 1 36 +2 1 18+......36 1 36 = Count yourself.

2 is a binomial distribution satisfying b (n, 1 2) then it is expected to be =3 1 2=

An event is known to have a probability of occurrence of p and a probability of non-occurrence of 1-p.

When an event occurs, a score is A; When an event does not occur, a score of be(x) is scored

ap + b(1-p)

This is a mathematical expectation problem for a discrete and immediate variable function:

According to the desired formula, there is e(x)=x*p(x).

In the same way: e(y)= (y*p(y)) = (y*p(x)) here: p(y)=p(x) because x and y are monotonic here: y=t(1-e -ax).

Here: x obeys a Poisson distribution with the parameter of , i.e., p(x)=(x)(e - x!).

Here: x takes 0, 1, 2....

1. Calculate e(x)=2 3 first; recalculate e(x2)=1 2; Calculate d(x)=e(x2)-[e(x)]2=1 2-4 9=1 18;

e(y)=2/3,d(y)=1/900

2. Use the central limit theorem when calculating probability p.

That's what I think. I didn't say how much it cost to trigger d, so I thought it was 0

If the value is too large, it will be reduced first, A will trigger the cost of 2, B will trigger the cost of 4, and C will trigger the cost of 6, a total of 1000.

1.Set up event a: A allows B and allows C and allows D and triggers D

The probability of this is p(a)=50%*50%*50%=1 8, and the total cost is 2+4+6=12;

2.Set up event B: A allows B and allows C but not D

This probability p(b) = 50% * 50% * 50% = 1 8, a total of 2 + 4 + 6 = 12;

3.Set up event C: A allows B but not C

Probability p(c) = 50% * 50% = 1 4, a total of 2 + 4 = 6;

4.Set up event D: A did not allow B

probability p(d) = 50% = 1 2, a total of 2 is consumed;

The so-called B triggers C, and then A retriggers B, which is equivalent to A allowing B but not allowing C.

So these special cases are actually included in the above four events.

So, each of the above events is a round, and we assume that a total of n rounds occurred in these 1000 points, regardless of which ABCD occurred.

Then A happened n 8 times, B happened n 8 times, C happened n 4 times, and D happened n 2 times.

Yes: 12*n 8+12*n 8+6*n 4+2*n 2=1000 The expectation for the number of occurrences n is n=2000 11

Then the only event A that can trigger d is the number of occurrences of n 8=250 11

So the mathematical expectation of the number of departures d should be 250 11

Personal thoughts, if it helps, I hope to [choose as a satisfactory answer], if you have any questions, please discuss together, thank you.

Let the number of triggers of A be x, then the number of times B is x 2, the number of times C is x 4, and the number of times D is x 8.

2000x+4000*x/2+6000*x/4《10000005500x《1000000

x "d of the number of times.

The number of triggers for integer d is 22.

There are a total of 4 3 2 1 = 24 equal probability conjunctions. Only 0 points are possible. 3 points. 6 points. 12 points and four situations:

12 points: 1 kind.

6 points: 4 3 2 = 6 types.

3 points: 4 2 = 8 kinds.

0 points: 3 (1 1 1 + 2 1) = 9 types.

Then the probability of no less than 6 points p=(6+8+9) 24=23 24

Question 1, [in the process of calculation, copy a=1 (2 )] 1), x n(0,1), y n(0,1), and x and y are independent of each other, and the joint distribution density function of x and y f(x,y)=a e (-x 2-y 2).

e[x²/(x²+y²)]=∫(-x²f(x,y)dxdy/(x²+y²)=a²∫(x²e^(-x²/2-y²/2)dxdy/(x²+y²)。

Let x= cos , y= sin , e[x (x +y )]=a (0,2 )cos d (0, )e (- 2)d =1 2.

x p(1), its probability distribution function p(x=k)=(1 e) (k!).)，e(x)=d(x)=1。∴e(x²)=d(x)+e(x)=2。

p[x=e(x²)]=p(x=2)=1/(2e)。

FYI.