Suppose the random variables X, Y, and z have joint distribution as follows: f(x, y, z) = xy^2 z/180, x = 1, 2, 3; y = 1, 2; z = 1, 2, 3 Find the two-dimensional marginal distributions f_1, 2(x, y), f_1, 3(x, z) and f_2, 3(y, z). Find the marginal distributions f_1(x), f_2(y), and f_3(z). Find P(Y = 2|X = 1, Z = 3). Find P(X greaterthanorequalto 2, Y = 2|Z = 2). Are X, Y, and Z independent?

Multiple discrete random variables make up the joint distribution function.
A joint probability of a given random variables is computed using it.
The joint function can be used to get the marginal function of a random variables.

Part a: f(x, y, z) = xy^2 z/180, x = 1, 2, 3; y = 1, 2; z = 1, 2, 3 .

f(x, y) = ∑(z = 1 →3) f(x, y, z)

= ∑(z = 1 →3) xy^2 z/180

= xy^2 /180 ( 1 + 2 + 3)

= xy^2/30

f(x, z) = ∑(y = 1 →2) f(x, y, z)

= ∑(y = 1 →2) xy^2 z/180

= xz /180 ( 1² + 2²)

= xz/36

f(y, z) = ∑(x = 1 →3) f(x, y, z)

= ∑(x = 1 →3) xy^2 z/180

= zy^2/180 (1 + 2 + 3)

= zy^2/30

Part b: marginal distributions f_1(x), f_2(y), and f_3(z).

f(x) = ∑(y = 1 →2) f(x, y)

= ∑(y = 1 →2) xy^2/30

= x/6

f(y) = ∑(x = 1 →3) f(x, y)

= ∑(x = 1 →3) xy^2/30

= y^2/5

f(z) = ∑(x = 1 →3) f(x, z)

= ∑(x = 1 →3) xz/36

= z/6

Part c: P(Y = 2|X = 1, Z = 3)

P(Y = 2|X = 1, Z = 3) = (1 x 2 x 3)/180 / (1 x 3)/36

P(Y = 2|X = 1, Z = 3) = 0.8

Part d: P(X ≥ 2, Y = 2|Z = 2)

P(X ≥ 2, Y = 2|Z = 2) = P(2, 2, 2)+P(3,2,2)/P(Z = 2)

P(X ≥ 2, Y = 2|Z = 2) = [(2 x 2² x 2)/180 + (3 x 2² x 2)/180] / (2/6)

P(X ≥ 2, Y = 2|Z = 2) = 0.667

Part e: In this case, X, Y, and Z all independent variables since their joint function equals the sum of their marginal functions.

f(x, y, z) = f(x) x f(y) x f(z)

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