Answer:
a) To find the probability that X1 is less than or equal to 0.5 and X2 is greater than or equal to 0.25, we need to integrate the given density function over the region where X1 ≤ 0.5 and X2 ≥ 0.25.
P(X1 ≤ 0.5, X2 ≥ 0.25) = ∫∫(x1,x2) f(x1,x2) dxdy
where the limits of integration are:
0.25 ≤ x2 ≤ 1
0 ≤ x1 ≤ 0.5
Substituting the given density function:
P(X1 ≤ 0.5, X2 ≥ 0.25) = ∫0.25^1 ∫0^0.5 (x1 + x2) dx1 dx2
Evaluating the inner integral:
P(X1 ≤ 0.5, X2 ≥ 0.25) = ∫0.25^1 [(x1^2/2) + x1x2] |0 to 0.5 dx2
Simplifying the expression:
P(X1 ≤ 0.5, X2 ≥ 0.25) = ∫0.25^1 [(0.125 + 0.25x2)] dx2
Evaluating the upper and lower limits:
P(X1 ≤ 0.5, X2 ≥ 0.25) = [0.125x2 + 0.125x2^2] |0.25 to 1
Substituting the limits:
P(X1 ≤ 0.5, X2 ≥ 0.25) = [(0.125 + 0.125) - (0.03125 + 0.015625)]
Solving for the final answer:
P(X1 ≤ 0.5, X2 ≥ 0.25) = 21/64
Therefore, the probability that X1 is less than or equal to 0.5 and X2 is greater than or equal to 0.25 is 21/64.
b) To find the probability that X1 + X2 is less than or equal to 1, we need to integrate the given density function over the region where X1 + X2 ≤ 1.
P(X1 + X2 ≤ 1) = ∫∫(x1,x2) f(x1,x2) dxdy
where the limits of integration are:
0 ≤ x1 ≤ 1
0 ≤ x2 ≤ 1-x1
Substituting the given density function:
P(X1 + X2 ≤ 1) = ∫0^1 ∫0^(1-x1) (x1 + x2) dx2 dx1
Evaluating the inner integral:
P(X1 + X2 ≤ 1) = ∫0^1 [(x1x2 + 0.5x2^2)] |0 to (1-x1) dx1
Simplifying the expression:
P(X1 + X2 ≤ 1) = ∫0^1 [(x1 - x1^2)/2 + (1-x1)^3/6] dx1
Evaluating the integral:
P(X1 + X2 ≤ 1) = [x1^2/4 - x1^3/6 - (1-x1)^4/24] |0 to 1
Substituting the limits:
P(X1 + X2 ≤ 1) = (1/4 - 1/6 - 1/24) - (0/4 - 0/6 - 1/24)
Solving for the final answer:
P(X1 + X2 ≤ 1) = 1/8
Therefore, the probability that X1 + X2 is less than or equal to 1 is 1/8.