Let Lx denote the event that the student finishes the exam in less than x hours and has probability x/2 and by taking x =1 it is 1/2 , 0 <= x <= 1, and F is the event that the student uses the full hour, which means that F is also the event that he is not finished in less than an hour.
P(F) = $P(L_{1}^{c})$ = 1−$P(L_{1})$ = 1 - 1/2 = 0.5
Now, the event that the student is still working at time .75 is the complement of the event $L_{0.75}$, or we can say that Now, we are looking for the probability that the full hour is used, given that the student is still working after 0.75 hours, so the desired probability is obtained from
P( F | $L_{0.75}^{c}$ ) = P( F ∩ $L_{0.75}^{c}$ ) / $P(L_{0.75}^{c})$
=P(F) / 1 − $P(L_{0.75})$
=(1/2) / (1 - (0.75/2))
=0.50 / .625
=0.8
