Recent content by JP16

We use the mean as 1/3. E(XY) = E(X) E(Y) since X and Y are independent. So than it is equal to the Var(X). Thanks for both of your help. Much appreciated. :)

Thank you for confirming. Although this problem is solved, I would like to know more. Before I was trying to Cov(X,Z) directly without subbing the 'X+Y', and problems rose there. Because Cov(X,Z) = E(XZ) - E(X)E(Z) then, E(XZ) = ∫Dxz fX,Z(x, z) dD, where fX,Z(x, z) is the prob. density...

Homework Statement Let X ~ Exponential(3) and Y ~ Poisson(5). Assume X and Y are independent. Let Z = X + Y. Compute the Cov(X,Z).Homework Equations I know Cov(X, Z) = E(XZ) - E(X)E(Z). But how do I compute E(XZ) and E(Z) ?? Since for E(XZ), I would need the pdf/pmf (Exp is abs cts, while...

Thanks for the fast responses and guiding me in the right direction. Any guidance for the other question?

Oh my bad, it should just be (2/4) * (1/6) + (1/4)*(1/6). (2/4) : probability of having 2 heads in the total # of outcomes of flipping 2 coins. (1/6) : probability of rolling a 1. (1/4) : probability of both coins being heads. (1/6) : probability of rolling a 2. Am I making any...

As HallsofIvy said the possible outcomes, the set would contain {TH1,HT1, HH2}. Then the probability of that would just be, [SIZE="4"]2(\frac{2}{4}\frac{1}{6})+\frac{1}{4}\frac{1}{6} = \frac{5}{24} Would this be correct? OH, that makes sense. Then, P(1st die 2 | least 3 dice show 2)...

Thank for the quick reply. For 1) Denote by A, the # of heads. by B the # on dice. Then P(B|A) = P(B and A)/ P (A) - For P(A), for at least 1 heads the prob. is 3/4. For intersection, possible sample space is for dice to be either 1 or 2 (1/3). But I'm confused as to what is the...

Homework Statement 1) S'pose we flip 2 fair coins, and roll one fair 6 sided die. What is the probability that the number of heads equals the number showing on the die? 2) We roll 4 fair six sided dice. What is the conditional prob. that the first die shows 2, conditional on the event that at...

oh i tried doing that with 2x=y, now i see why that would be wrong. c3 = -2c2 c = -2 => not valid since c has to be > 0, since its in the 'same direction'. moving on. c4= 4c2 c = 2 => which holds true so then a = 4. I know you need to solve equations, but for questions such as b and...

Homework Statement The question states: "find all scalars c, if any exist such that given statement is true." It also suggests trying to do without pencil and paper. a) vector [c2, c3, c4] is parallel to [1,-2,4] with same direction. b) vector [13, -15] is a linear combination of vectors...

i kind of realize that now, but i just couldn't see it with the summations. no, i still don't know where to go afterwards. what do you mean by "what are"? sorry, i am just as much frustrated.

oh it is soo much clear after writing it in factorial way. so: [SIZE="4"] \frac{(-1)^jn!}{j!(n-j)!} - \frac{n!}{j!(n-j)!} = \frac{n!(-1^j - 1)}{j!(n-j)!} where [(-1)^j -1] = -2 since j is odd. and so yes, then it is (-2)*C(n,j)

ohh then (-2)*C(n,j) = (-2)*(1+1)^n ?

how did you get the -2? (1+1)^n = C(n,j) = 2^n (1-1)^n = (-1)^j*C(n,j) = 0 ah! this is really confusing.

oh yes, i was overlooking the subtraction, and thinking of sum, hence the even terms. But yes they would only give the odd terms. How would it give me it twice though && how do i go about simplifying? Let's say:#of_term...(1+1)^n...(1-1)^n....difference 0......1......0......1 what would be the...