Let X and Y be continuous random variables with joint density function
f(x, y) = 24xy, 0<x<1, 0<y<1-x
calculate p(y<x | x=1/3)

我的解法：
先算marginal f(y) = 12y(1-y)^2
p(y<x | x=1/3)  = p(y<1/3) = integrage f(y) from 0 to 1/3 = 2/27

请问这么算怎么错了？  答案是1/4