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!
Best Answer
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 $P(L_{0.75}^c)=1-\frac{0.75}{2}=0.625$