Hi all, The question is: Given that X follows a geometric distribution, prove that P(X>j+k | X>k) = P(X>j). So this is all I have: P(X>k+j) = (1-p)^k+j, (a proof covered in the text) = (1-p)^k • (1-p)^j, No, what I want to say is, since P(X>k) = (1-p)^k = 1, then P(X>k+j) = 1 • (1-p)^j, = P(x>j). But I don't know if I'm allowed to separate the expression (1-p)^k • (1-p)^j into separate probabilities... Thanks in advance!!!