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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
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