Basic problem on conditional probability. Need help in understanding the solution I am confused about below basic problem of conditional probability. Here is the problem and the given solution.
A student is taking a one-hour-time-limit makeup examination. Suppose the probability that the student will finish the exam in less than $x$ hours is $\frac{x}2, \forall 0 \leq x \leq 1$. Then, given that the student is still working after $.75$ hour, what is the conditional probability that the full hour is used?
Solution. Let $L_x$ denote the event that the student finishes the exam in less than $x$ hours, $0 \leq x \leq 1$, and let $F$ be the event that the student uses the full hour. Because $F$ is the event that the student is not finished in less than $1$ hour,
$P(F) = P(L_1^c) = 1 - P(L_1) = 0.5$
Now, the event that the student is still working at time .75 is the complement of the
event $L_.75$, so the desired probability is obtained from
$$ P(F|L_.75^c) = \frac {P(F\color{blue}{\cap}L_.75^c)} {P(L_.75^c)}$$
$$= \frac {P(F)} {1 - P(L_.75)} $$
$$= \frac {0.5}{0.625} = 0.8 $$
QUESTION:
I do not understand only the bottom expression. How did we get $P(F)$ from $P(FL_.75^c)$ ?
And how did we compute $P(L_.75^c)$ so we got 0.625?
Thank you!
 A: Note that $FL_{.75}^c$ (a full hour is used and the student is not ready in less than $.75$ hours) is exactly the same event as $F$ (a full hour is used). You could say that $F\subset L_{.75}^c$ and consequently $F\cap L_{.75}^c=F$.
$P(L_x^c)=1-\frac{x}{2}$ leads directly to  $P(L_{0.75}^c)=1-\frac{0.75}{2}=0.625$
A: The event that you use more than $0.75$ of the hour and use the full hour is logically equivalent to just using the full hour. So
$$P(F, L_{.75}^c) = P(\text{Uses full hour and uses  0.75 of the hour or more}) = P(F).$$
And $1-P(L_{.75})=1-0.75/2$ by simply using the formula your were given for the probability.
A: *

*$F=F\cap L_{.75}^c$

*$P(L_{.75}^c)=1-P(L_{.75})$ and $P(L_x)=x/2$
A: How about we use Venn Diagram to solve this problem
Area of Venn Diagram $= 1$
Area of Event $A$ i.e. $1$ hour to complete $=\frac12 \times 1=0.5$
Area of Event $B$ i.e. $\frac34$ hour to complete $=\frac12 \times \frac34= \frac38$
then Area $B'=1-\frac38=\frac58$
Area of $A' = 1-0.5=0.5$
Note: event $B$ is comprised/subset of Event $A$
So $P(A∩ \cap B')$ is area outside area of Event $A=1-0.5=0.5$
$P(A|B')=\frac{P(A \cap B')}{P(B')}=\frac{1/2}{5/8}=0.8$
Please correct me in case I am wrong. :) 
