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这道题就成这个问题
x1 and x2 follows uniform distribution U[0,1]
what is the probability that |x1-x2|>1/3?
case a:
0<=x1<1/3
then as long as x2<=x1+(1/3), |x1-x2|>1/3
so the probability is x1+(1/3)
case b:
1/3<=x1<2/3
then as long as x1-(1/3)<=x2<=x1+(1/3)
so the probability is 2/3
case c:
2/3<=x1<=1
similar to case a
probability is (4/3)-x1
then integrate all the probability, we get 5/9
if we use area method, consider (x1, x2) as a point in square [0，1]^2
then |x1-x2|<1/3 is bounded by two lines, x1-x2<1/3 and x2-x1<1/3
by subtracting the two triangles, the area between the lines is 5/9
done..
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