Die is rolled 3 times and 1 more

  • Thread starter RsMath
  • Start date
  • #1
7
0
I have these two question I hope someone could check them out for me :

1) A die is rolled 3 times, let x be the sum of the 3 results .

what is var(x) ?

I started by calculating E(x)= Segma(1 to 3 (E(Xi)) when Xi is the number we got in every throw .
E(Xi)=3.5
then E(x) = 10.5 (Expected of the sum)

now what should I do to find the var of x ?

2) we have n balls and n^2 chains , we throw the balls randomly into the chains.
we pick y randomly - a chain , what's the probability that it has exactly 3 balls in it .

ok, I've kinda solve this question but i'm not sure if the answer is correct .. we choose 3 balls from the n balls , and we do it in n!/((n-3)!*3!) and we calculate every ball by the probability that it goes into y , 1/n^2 .. thus final answer is :

n!
-------------
(n-3)!*3!*n^6

can you help me complete the first Q and check the second one for me.

thanks.
 

Answers and Replies

  • #2
85
1
Let x be a random variable that stands for a single toss of the die.

x p[x]
1 1/6
2 1/6
......
6 1/6

Then E[x] = 3.5 and E[x^2] = 15.1667 and var(x) = 2.9167

We then have Y = X + X + X. Thus,

[tex] \rho^2_Y = \frac{\rho^2_X}{N}= \frac{2.9167}{3}[/tex]
 
Last edited:
  • #3
7
0
thanks tedbradly.
you have found the variance of the mean by dividing by 3 .
shouldn't it be :
var(x+x+x) = var(x)+var(x+x)+2*cov(x,x+x) =
var(x)+4*var(x)+2*(E(x(x+1))-E(x)*E(x+x) = 9*var(x) ?? and not var(x)/3 ?
I just need to make sure .
 
  • #4
85
1
thanks tedbradly.
you have found the variance of the mean by dividing by 3 .
shouldn't it be :
var(x+x+x) = var(x)+var(x+x)+2*cov(x,x+x) =
var(x)+4*var(x)+2*(E(x(x+1))-E(x)*E(x+x) = 9*var(x) ?? and not var(x)/3 ?
I just need to make sure .

I have no idea what all that is that you typed, but I just used the central limit theorem.
 
  • #5
2,044
312
I have no idea what all that is that you typed, but I just used the central limit theorem.

he's using var(X+Y) = var(X) + var(Y) + 2cov(X,Y).

since X and Y are independent the covariance is 0, so you get var(X+Y) = var(X) + var(Y) .

likewise var(X+Y+Z) = var(X)+var(Y+Z) = var(X)+var(Y)+var(Z) (if X, Y and Z are independent.)

If X,Y and Z have an identical distribution then: var(X+Y+Z) = 3var(X)
 
  • #6
85
1
he's using var(X+Y) = var(X) + var(Y) + 2cov(X,Y).

since X and Y are independent the covariance is 0, so you get var(X+Y) = var(X) + var(Y) .

likewise var(X+Y+Z) = var(X)+var(Y+Z) = var(X)+var(Y)+var(Z) (if X, Y and Z are independent.)

If X,Y and Z have an identical distribution then: var(X+Y+Z) = 3var(X)
Oh. I was doing variance of the mean, which is wrong.
 

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