Let X and Y be independent Bernoulli random variables with parameter 1/2, and let Z be the random variable that returns the remainder of the division of X + Y by 2.

(a) Prove that Z is also a Bernoulli random variable, also with parameter 1/2.
(b) Prove that X; Y;Z are pairwise independent but not mutually independent.
(c) By computing Var[X+Y +Z] according to the alternative formula for variance and using the variance of Bernoulli r. v.'s, verify that Var[X +Y +Z] =Var[X]+Var[Y ]+Var[Z]

** Kindly use the properties of Bernoulli random variables for solving questions. Thanks!

a) Since Z is a remainder of a divition by 2, then it can only take the values 0 and 1, thus it is a bernoulli random variable. Z is 1 when either X or Y is 1 and the other is 0, therefore

P(Z=1) = P(X = 1 and Y = 0) + P(X = 0 and Y=1)

For independence of X and Y , that sum is equal to

P(X=1)*P(Y=0) + P(X=0) *P(Y=1) = 1/2*1/2 + 1/2*1/2 0 1/4 + 1/4 = 1/2.

The probability for success is 1/2, so effectively Z is a bernoulli with parameter 1/2.

b) X and Y are independent because we are given that information.

If we prove that X and Z are independent, then we can prove with a similar argument that Y and Z are.

Lets compute every possible probability. Note that since both X and Z are bernoullis with parameter 1/2, then P(X=k)*P(Z = j) = 1/2*1/2 = 1/4 for k,j in {0,1}.

P(X = 0 and Z = 0): if X is 0, then Z is 0 only if Y is 0 (otherwise it would be 1), thus

P(X = 0 and Z = 0) = P(X=0 and Y = 0) = P(X=0) *P(Y=0) (independence) = 1/2*1/2 = 1/4

P(X=0, Z=1) = (for Z to be 1 given than X is 0, we need Y to be 1) P(X=0 and Y=1) = P(X=0) *P(Y=1) = 1/2*1/2 = 1/4

P(X=1, Z=0) = P(X=1, Y=1) = P(X=1)*P(Y=1) = 1/4 (If X is 1, then Z is 0 when Y is also 1)

P(X=1,Z=1) = P(X=1,Y=0) = P(X=1)*P(Y=0) = 1/4 (If X is 1, then Z is 1 only if Y is 0)

This show that X and Z are independent (hence so are Y and Z)

However, Z is heavily dependant on X and Y, for example, If X and Y are both 1, then Z should be 0, this means that

P(X=1,Y=1,Z=1) = 0

but P(X=1)*P(Y=1)*P(Z=1) = 1/2*1/2*1/2 = 1/8 ≠ 0

Thus, X, Y and Z are pairwise independent but not mutually independent.

c)

Since X is bernoully with parameter 1/2, then

V(X) = 1/2-(1/2)² = 1/4

Therefore,

V(X) + V(Y) + V(Z) = 1/4*3 = 3/4

On the other hand,

V(X+Y+Z) = E( ( X + Y + Z )² ) + E(X+Y+Z)²

X+Y+Z could take any value from 0 to 3. since X, Y and Z cant take the value 1 at the same time, then X+Y+Z cant be 3. It cant also be 1, because if X and Y are both 0, then so is Z, and if X is 1 and Y is 0 (or otherwise), then Z is 1.

Thus, X+Y+Z can take the values 0 and 2. It only takes the value 0 when X, Y and Z is 0, and that happens only if X and Y are 0. Therefore

P(X+Y+Z = 0) = P(X = 0, Y=0, Z=0) = P(X=0,Y=0) = P(X=0)*P(Y=0) (independence) = 1/2*1/2 = 1/4.

and, as a result,

P(X+Y+Z=2) = 1-1/4 = 3/4.

As a consequence,

E(X+Y+Z)²= (0*1/4+3/4*2)² = (3/2)² = 9/4

On the other hand, (X+Y+Z)² can take the values 0, with probability 1/4 (When X+Y is 0), and 2² = 4, with probability 3/4 (When X+Y is 2)

Thus, E((X+Y+Z)²) = 0*1/4+4*3/4 = 3

We conclude that

V(X+Y+Z) = 3-9/4 = 3/4 = V(X)+V(Y)+V(Z)

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

ask for details followreport by buttfacelily 2 minutes ago

answers

rauschg

rauschg · hand

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step-by-step explanation:

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.