conditional probability that a freshman will graduate

Question

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?

Answer 1

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 $P(A|B^c)$.

$$ \begin{align} P(A|B^c) &\overset{\text{Bayes}}{=} \frac{P(B^c|A)P(A)}{P(B^c)} \\ &= \frac{\left(1-P(B|A)\right)P(A)}{1-P(B)} \\ &= \frac{\left(1-\frac{3}{4}\right)\frac{1}{2}}{1-\frac{5}{8}} \\ &= \frac{1/8}{3/8} \\ &= \frac{1}{3} \end{align} $$