If the joint probability distribution of X and Y f(x, y) = (x + y)/2

If the joint probability distribution of X and Y f(x, y) = (x + y)/2, x=0,1,2,3; y=0,1,2, Compute the following a. P(X≤2,Y =1) b. P(X>2,Y ≤1) c. P(X>Y) d. P(X+Y=4)

Solutions

Expert Solution

We are given here that:
f(x, y) = (x + y)/2

The non normalized probability distribution of X , Y here is computed as:

	x = 0	x = 1	x = 2	x = 3
y = 0	0	1/2 = 0.5	2/2 = 1	3/2 = 1.5
y = 1	1/2 = 0.5	2/2 = 1	3/2 = 1.5	4/2 = 2
y = 2	2/2 = 1	3/2 = 1.5	4/2 = 2	5/2 = 2.5

Sum of all cells above:
= 0.5 + 1 + 1.5 + 0.5 + 1 + 1.5 + 2 + 1 + 1.5 + 2 + 2.5 = 15

Therefore the normalized PMF for (X, Y) here is given as:

The probabilities now are computed as:

a) P(X <= 2, Y = 1) = f(0,1) + f(1,1) + f(2,1) = 0.5 + 1 + 1.5 = 3
Therefore 3/15 = 0.2 is the required probability here.

b) P(X > 2, Y <= 1) = f(3, 0) + f(3, 1) = 1.5 + 2 = 3.5
Therefore 3.5/15 = 0.2333 is the required probability here.

c) P(X > Y) = f(1,0) + f(2,0) + f(2,1) + f(3,0) + f(3,1) + f(3,2)
= 0.5 + 1 + 1.5 + 1.5 + 2 + 2.5 = 9
Therefore 9/15 = 0.6 is the required probability here.

d) P(X + Y = 4) = f(3, 1) + f(2, 2) = 2 + 2 = 4
Therefore 4/15 = 0.2667 is the required probability here.

orchestra answered 1 year ago

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