Answer:
P(X, Y) = 1/(20.398 ) if X ∈ (2, 8.2) and Y ∈ (3.01, 6.3)
P(X, Y) = 0 if X ∉ (2, 8.2) and Y ∉ (3.01, 6.3)
Step-by-step explanation:
We know that the possible values of X (all with the same probability) are:
2 < X < 8.2
And for Y we have the range (also uniform):
3.01 < Y < 6.3
Then we have a rectangle of area.
(8.2 - 2)*(6.3 - 3.01) = 20.398
Such that the probability for all the points inside the rectangle should be exactly equal, then we will have:
P(X, Y) = 1/(20.398 ) if X ∈ (2, 8.2) and Y ∈ (3.01, 6.3)
Such that if we integrate this in the given region, the integral will be equal to 1, as we should expect.
And zero if the point is outside that region, this is:
P(X, Y) = 0 if X ∉ (2, 8.2) and Y ∉ (3.01, 6.3)
