SOLUTION: Please help. Any help would be appreciated. Thank you.
We throw one fair coin and one fair die (they are indipendent) and we define the Random Variables X and Y so that the foll
Question 860902: Please help. Any help would be appreciated. Thank you.
We throw one fair coin and one fair die (they are indipendent) and we define the Random Variables X and Y so that the following outcomes are described: X=1 if we get H(Heads), and X=0 if we get T(Tails) and Y=outcome of the die. We also define the Random Variable Z = XY.
A) Define the sample space S and the Random Variables X and Y as functions of S.
B) Find the set of Rnadom Variable Z and count its mass of probability.
C) Find the mean value (E(Z)) and dispersion (Var(Z)) of Random Variable Z. Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website! A)Student says that the die is a 6 sided die
our sample space consists of pairs (a, b) where a is 0 or 1 depending on the coin toss (1 = heads, 0 = tails) and b is 1, 2, 3, 4, 5, 6 depending on die toss, we assume the die has 6 sides and each side is numbered
here is the sample space
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6)
there are 12 possible outcomes when throwing the coin and dice
S(X) is 1 if coin toss is heads
S(X) is 0 if coin toss is tails
S(Y) is i if die toss is i, where i is 1, 2, 3, 4 5, 6
B) sample space for Z when Z = XY
Z value probability
1 1/12
2 1/12
3 1/12
4 1/12
5 1/12
6 1/12
0 6/12 = 1/2
adding up the probabilities we get 1/2 + 1/2 = 1
C) E(z) = 0*(1/2) + 1*(1/12) + 2*(1/12) + ...+6*(1/12)
E(z) = 0 + (1/12)*(1 + 2 + 3 + 4 + 5 + 6)
E(z) = (1/12) * 21 = 21/12 = 7/4 = 1.75
Var(Z) = (0 - 21/12)^2 * (1/2) + (1 - 21/12)^2 * (1/12) + (2 - 21/12)^2 * (1/12) + ... + (6 - 21/12)^2 * (1/12)
Var(Z) = (0 - 1.75)^2 * (1/2) + (1/12) * ( (1 - 1.75)^2 + (2 - 1.75)^2 + (3 - 1.75)^2 + (4 - 1.75)^2 + (5 - 1.75)^2 + (6 - 1.75)^2 )
Var(Z) = 1.53125 + (1/12) * 35.875 = 4.52