In: Statistics and Probability
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)
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.