Yes this is wrong way,i dont know what i did, please find the correct one:::::::

at K=1 waiting time of station A=(0,1)*p.t.

at K=1 waiting time of station B=(0,1)*p.t.

For the second time, Since st.A wins the race(means A has sent data safely) so it will transfer new frame and for the second time station B will continue with old frame(it has not transferred old data)

So for A, waiting time=(0 to (2^1)-1)*p.t.

Station A transferring again after successful transferring at k=1, So we will start backoff algorithm for this frame with fresh parameter,so K=1 because its transferring new frame

That means waiting time of station of A ={0,1*p.t.}

For station B waiting time would be (0 to (2^2)-1)*p.t. here k=2 because count for THIS frame(old frame) become 2 since old backoff algo is not ended.

That means waiting time of station B {0, 1*p.t, 2*p.t, 3*p.t.}

You can see there are 8 possible pair out which 5 would be favourable case for A to win the race

Pair in which A wins i.e (A,B)={. (0,1*p.t),(0.,2*p.t,), (0,3*p.t ),(1*p.t,2*p.t),(1*p.t.,3*p.t) }

So 5/8=.625

Hope you Will get my approach.