We have two fair three-sided dice, indexed by i=1,2. Each die has sides labelled 1, 2, and 3. We roll the two dice independently, one roll for each die. For i=1,2, let the random variable Xi represent the result of the ith die, so that Xi is uniformly distributed over the set {1,2,3}. Define X=X2−X1.

Calculate the numerical values of following probabilities, as well as the expected value and variance of X:

P(X=0)= 1/3

P(X=1)= 2/9
P(X=−2)= 1/9
P(X=3)= 0
E[X]= 0
var(X)= 4/3

P(Y=0)= 1/3
P(Y=1)= 4/9
P(Y=2)= 0

Plz answer...

Calculate the numerical values of following probabilities, as well as the expected value and variance of X:

P(X=0)=
P(X=1)=
P(X=−2)=
P(X=3)=
E[X]=
var(X)=

Let Y=X2.
Calculate the following probabilities:

P(Y=0)=
P(Y=1)=
P(Y=2)=

Oh, dice probabilities! Finally, a topic that rolls right into my wheelhouse! Whether it's "dicey" business or not, here we go.

First, let's find the probability of each outcome for X.

If X2 is greater than X1, we get X = X2 - X1 = (1, 2) = -1.
If X2 is equal to X1, we get X = X2 - X1 = (1, 1) = 0.
If X2 is less than X1, we get X = X2 - X1 = (2, 1) or (3, 1) or (3, 2) = 1.

Since we have two fair three-sided dice, the probability of each outcome is 1/3.

The probability of X = -1 is 1/3 * 1/3 = 1/9.
The probability of X = 0 is 1/3 * 1/3 = 1/9.
The probability of X = 1 is 1/3 * 2/3 = 2/9.

To calculate the expected value, we multiply each outcome by its probability and sum them up:

Expected value of X = (-1) * (1/9) + 0 * (1/9) + 1 * (2/9) = 0.

Now, let's calculate the variance of X. The variance is the mean of the squared differences from the expected value.

Variance of X = ((-1 - 0)^2 * (1/9)) + ((0 - 0)^2 * (1/9)) + ((1 - 0)^2 * (2/9))
= (1/9) + 0 + (2/9)
= 3/9
= 1/3.

So there you have it! The probabilities are 1/9 for X = -1, 1/9 for X = 0, and 2/9 for X = 1. The expected value is 0, and the variance is 1/3. Keep rolling those dice, and remember, life is just a roll of the dice!

To calculate the probabilities, expected value, and variance of X, we need to analyze the possible outcomes and their corresponding probabilities.

Let's start by writing down all the possible outcomes of the two dice rolls. Since each die can have one of three numbers, there are a total of 3 * 3 = 9 possible outcomes. We can represent each outcome as (X1, X2), where X1 is the result of the first die and X2 is the result of the second die:

(1, 1), (1, 2), (1, 3),
(2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (3, 3).

Now, let's calculate the probabilities of each outcome. Since the dice are fair, each outcome is equally likely, so the probability of each outcome is 1/9.

Next, we need to define the random variable X = X2 - X1. We can calculate the value of X for each outcome:

X(1, 1) = 1 - 1 = 0,
X(1, 2) = 2 - 1 = 1,
X(1, 3) = 3 - 1 = 2,
X(2, 1) = 1 - 2 = -1,
X(2, 2) = 2 - 2 = 0,
X(2, 3) = 3 - 2 = 1,
X(3, 1) = 1 - 3 = -2,
X(3, 2) = 2 - 3 = -1,
X(3, 3) = 3 - 3 = 0.

Now we can calculate the probabilities of each value of X:

P(X = -2) = P((3, 1)) = 1/9,
P(X = -1) = P((2, 1), (3, 2)) = 2/9,
P(X = 0) = P((1, 1), (2, 2), (3, 3)) = 3/9 = 1/3,
P(X = 1) = P((1, 2), (2, 3)) = 2/9,
P(X = 2) = P((1, 3)) = 1/9.

Now we can calculate the expected value of X:

E[X] = (-2) * P(X = -2) + (-1) * P(X = -1) + 0 * P(X = 0) + 1 * P(X = 1) + 2 * P(X = 2).

Substituting the probabilities we calculated, we get:

E[X] = (-2) * (1/9) + (-1) * (2/9) + 0 * (1/3) + 1 * (2/9) + 2 * (1/9) = -1/9.

To calculate the variance of X, we need to calculate the deviations from the mean and their squared values:

(x - E[X])^2 * P(X = x).

For each value of X, the deviations are:

(-2 - (-1/9))^2 = (-(17/9))^2 = 289/81,
(-1 - (-1/9))^2 = (-8/9)^2 = 64/81,
(0 - (-1/9))^2 = (1/9)^2 = 1/81,
(1 - (-1/9))^2 = (10/9)^2 = 100/81,
(2 - (-1/9))^2 = (19/9)^2 = 361/81.

Now we can calculate the variance:

Var[X] = (289/81) * (1/9) + (64/81) * (2/9) + (1/81) * (1/3) + (100/81) * (2/9) + (361/81) * (1/9).

Simplifying this expression, we get:

Var[X] = (289 + 128 + 1 + 200 + 361) / (9 * 81) = 979 / (9 * 81) = 979 / 729 ≈ 1.344.

Therefore, the probabilities, expected value, and variance of X are:

P(X = -2) ≈ 1/9,
P(X = -1) ≈ 2/9,
P(X = 0) = 1/3,
P(X = 1) ≈ 2/9,
P(X = 2) ≈ 1/9,

E[X] = -1/9,

Var[X] ≈ 1.344.