First Backoff Race (X(0,1) ,Y(0,1))possible combinations : (0,0) (0,1) (1,0) (1,1)
Here Y wins implies (1,0)
Second Backoff Race (x,Y) possible combinations: ((here x packet is getting in collision second time so for X(0,1,2,3) and Y(0,1))
Now there are 8 possibilities out of which 1 is where x wins i,e.,(0,1)
so ans is 1/8 = 0.125