finding the probability to get a diploma For getting a diploma a person needs to go to $3$ interviews at $3$ teachers: $A,B,C$.
In each interview a teacher can give a positive opinion or negative opinion.
The person will go to interview at each of the teachers and after that they ($A,B,C)$ will meet and compare their opinions about that person.


*

*The probability that teacher $A$ will give a positive opinion is $0.75$.

*The probability that teacher $B$ will agree with the opinion of teacher $A$ is $\frac{2}{3}$.

*The probability that teacher $C$ will agree with the opinion of teacher $A$ is $0.25$.


The person will get the diploma only if at least $2$ of the teachers gave him a positive opinion.

Given that teacher $B$ gave the person a positive opinion, what is the probability that the person didn't get the diploma?

My Attempt: i tried to find the probability for the person to get a diploma and i got $\frac{5}{8}$ but i don't know how to continue.
The answer is $\frac{1}{28}$ but i don't know why.
 A: Map the entire sample-space:


*

*$P(    {A}\wedge    {B}\wedge    {C})=\frac34\cdot\frac23\cdot\frac14$

*$P(    {A}\wedge    {B}\wedge\neg{C})=\frac34\cdot\frac23\cdot\frac34$

*$P(    {A}\wedge\neg{B}\wedge    {C})=\frac34\cdot\frac13\cdot\frac14$

*$P(    {A}\wedge\neg{B}\wedge\neg{C})=\frac34\cdot\frac13\cdot\frac34$

*$P(\neg{A}\wedge    {B}\wedge    {C})=\frac14\cdot\frac13\cdot\frac34$

*$P(\neg{A}\wedge    {B}\wedge\neg{C})=\frac14\cdot\frac13\cdot\frac14$

*$P(\neg{A}\wedge\neg{B}\wedge    {C})=\frac14\cdot\frac23\cdot\frac34$

*$P(\neg{A}\wedge\neg{B}\wedge\neg{C})=\frac14\cdot\frac23\cdot\frac14$


Sum up the probabilities of the given event (a positive opinion made by $B$):


*

*$P(    {A}\wedge    {B}\wedge    {C})=\frac34\cdot\frac23\cdot\frac14$

*$P(    {A}\wedge    {B}\wedge\neg{C})=\frac34\cdot\frac23\cdot\frac34$

*$P(\neg{A}\wedge    {B}\wedge    {C})=\frac14\cdot\frac13\cdot\frac34$

*$P(\neg{A}\wedge    {B}\wedge\neg{C})=\frac14\cdot\frac13\cdot\frac14$


Sum up the probabilities of the desired event (a positive opinion made only by $B$):


*

*$P(\neg{A}\wedge    {B}\wedge\neg{C})=\frac14\cdot\frac13\cdot\frac14$


Divide the latter by the former in order to calculate the conditional probability: $\dfrac{\frac{1}{48}}{\frac{28}{48}}=\frac{1}{28}$
A: Since you know that B gave a positive opinion, you need to know whether A gave a positive opinion as well, or if not, whether C gave a positive opinion. 
B agreeing with the opinion of A is the same as A agreeing with the opinion of B. So A has a $\frac 13$ probability of having given a negative opinion. 
If they gave a negative one ($\frac 13$ probability), then C will give a negative opinion as well $\frac 14$ of the time. 
Therefore, the probability that at the person had not the diploma is $\frac 13 \times \frac 14= \frac{1}{12}$.
A: You can use Bayes's Law symbolically, but it may be easier to just enumerate the possibilities in plain English.  $B$ gives a positive response when either $A$ gives a positive response and $B$ agrees (with probability $(3/4)(2/3) = 1/2$), or when $A$ gives a negative response and $B$ disagrees (with probability $(1/4)(1/3) = 1/12$).  The probability that $B$ is positive is therefore $1/2+1/12 = 7/12$.
On the other hand, with $B$ positive, the only way the student fails to receive the diploma is if $A$ is negative, $B$ disagrees, and $C$ agrees (with $A$).  That happens with probability $(1/4)(1/3)(1/4) = 1/48$.
Thus the probability that the student fails to receive a diploma, given that $B$ was positive, is $1/48$ divided by $7/12$, or $1/28$.
ETA: Symbolically:
$$
\begin{align}
P(\text{no diploma} \mid \text{$B$ positive})
    & = \frac{P(\text{$B$ positive, no diploma})}{P(\text{$B$ positive})} \\
    & = \frac{P(\text{$A$ negative, $B$ positive, $C$ negative})}
             {P(\text{$A$ positive, $B$ agrees with $A$}) +
              P(\text{$A$ negative, $B$ disagrees with $A$})} \\
    & = \frac{P(\text{$A$ negative, $B$ disagrees with $A$, $C$ agrees with $A$})}
             {P(\text{$A$ positive, $B$ agrees with $A$}) +
              P(\text{$A$ negative, $B$ disagrees with $A$})} \\
    & = \frac{(1/4)(1/3)(1/4)}{(3/4)(2/3)+(1/4)(1/3)} \\
    & = \frac{1/48}{7/12} = \frac{1}{28}
\end{align}
$$
