Solution:
There are two possibilities, A comes first, B comes first, B comes first, A comes first.
A and so on b:
There are two situations, A 0 ~ 3 hours to arrive, and 3 hours later.
The combined probabilities of these two cases are (3/4)*( 1/4)==3/ 16 and respectively.
Integrate 3 ~ 4 ((kloc-0//4) * ((4-t)/4) dt, and the result is 1/32.
So the probability of A and B waiting is 3/ 16+ 1/32==7/32.
B and so on a:
It is also two cases, B 0 ~ 2 hours to arrive, 2 hours to arrive.
They are: (1/2) * (1/2) =1/4 and.
The definite integral is 2 ~ 4 ((kloc-0//4) * ((4-t)/4) dt = =1/8.
Therefore, the probability of waiting for B and A is 3/8.
So the final probability is1/32+3/8 =19/32.