Probability of not preparing for exam Given that a student had prepared, the probability of passing a certain entrance
exam is $0.99$. Given that a student did not prepare, the probability of passing the
entrance exam is $0.05$. Assume that the probability of preparing is $0.7$. The
student fails in the exam. What is the probability that he or she did not prepare? 
 A: Refer to the probability tree diagram below:

We will find:
$$P(Pr'|Fail)=\frac{P(Fail\cap Pr')}{P(Fail\cap Pr')+P(Fail\cap Pr)}=\\
\frac{0.3\cdot 0.95}{0.3\cdot 0.95+0.7\cdot 0.01}=0.976.$$
A: Let $A$ be the event that a student prepares for the examination; let $B$ be the event that a student passes the examination.  We are given the following information:
\begin{align*}
\Pr(B \mid A) & = 0.99\\
\Pr(B \mid A^C) & = 0.05\\
\Pr(A) & = 0.7
\end{align*}
from which we can obtain the following information
\begin{align*}
\Pr(B^C \mid A) & = 1 - \Pr(B \mid A) = 1 - 0.99 = 0.01\\
\Pr(B^C \mid A^C) & = 1 - \Pr(B \mid A^C) = 1 - 0.05 = 0.95\\
\Pr(A^C) & = 1 - \Pr(A) = 1 - 0.7 = 0.3
\end{align*}
We wish to find the probability that a student has not prepared for the examination given that that student failed the examination, which is 
\begin{align*}
\Pr(A^C \mid B^C) & = \frac{\Pr(A^C \cap B^C)}{\Pr(B^C)}\\
                  & = \frac{\Pr(B^C \mid A^C)\Pr(A^C)}{\Pr(B^C \mid A)\Pr(A) + \Pr(B^C \mid A^C)\Pr(A^C)}\\
                  & = \frac{(0.95)(0.3)}{(0.01)(0.7) + (0.95)(0.3)}\\
                  & \approx 0.97602739726
\end{align*}
