conditional probability that a freshman will graduate 
A university finds that 75% of its graduating seniors scored above 80 on the entrance
  exam, while only 50% of those who fail to graduate score above 80. They also find that half
  of the entering freshmen graduate. What is the conditional probability that a freshman will
  graduate given that:
  (a) he or she scored above 80 on the entrance exam;
  (b) he or she scored 80 or below on the entrance exam?

My solution:
P(graduating seniors with score above 80)=$\frac{3}{4}$
P(non-graduating seniors with score above 80)=$\frac{1}{2}$
P(freshman will graduate)=$\frac{1}{2}$
(a)Using Bayes, $\frac{P(freshman will graduate, scored above 80)}{P(scored above 80)}$=
$\frac{1}{2}$ * $\frac{3}{4}$/$\frac{1}{2}$+$\frac{3}{4}$=$\frac{3}{10}$
(b)$\frac{1}{2}$
As given condition is either he scores above or below, it doesn't make any difference to the result.
Am I correct?
 A: We introduce the following notation.


*

*$\Omega$ - the set of admitted students, which we assume to be finite and non-empty.

*$A$ - the set of graduating students.

*$B$ - the set of students who scored above $80$ on the entrance exam.

*$P$ - the uniform probability on $\Omega$.


It is given that
$$
\begin{align}
P(B|A) &= \frac{3}{4} \\
P(B|A^c) &= \frac{1}{2} \\
P(A) &= \frac{1}{2}
\end{align}
$$
a. We are to calculate $P(A|B)$.
$$
\begin{align}
P(B) &\overset{\text{total prob.}}{=} P(B|A)P(A)+P(B|A^c)P(A^c) \\
&= \frac{3}{4}\frac{1}{2}+\frac{1}{2}\left(1-\frac{1}{2}\right) \\
&= \frac{5}{8} \\
P(A|B) &\overset{\text{Bayes}}{=} \frac{P(B|A)P(A)}{P(B)} \\
&= \frac{\frac{3}{4}\frac{1}{2}}{\frac{5}{8}} \\
&= \frac{3}{5}
\end{align}
$$
b. We are to calculate $P(A|B^c)$.
$$
\begin{align}
P(A|B^c) &\overset{\text{Bayes}}{=} \frac{P(B^c|A)P(A)}{P(B^c)} \\
&= \frac{\left(1-P(B|A)\right)P(A)}{1-P(B)} \\
&= \frac{\left(1-\frac{3}{4}\right)\frac{1}{2}}{1-\frac{5}{8}} \\
&= \frac{1/8}{3/8} \\
&= \frac{1}{3}
\end{align}
$$
