Suppose you're at a college campus. $3/4$ of the people on the campus are students or professors from that college, and the rest $1/4$ aren't. When asked a question, students and professors from that college will give you a correct answer every time, and those that aren't from the college will give you a correct answer $3/8$th of the time.
a) You stop a random person and ask for directions to place $A$ belonging on campus and he gives you an answer. What is the probability that the answer is true?
b) You ask the same person the same question again and he gives you the exact same answer. What is the probability that answer is correct now?
Solution a):
Using the total probability theorem, I make two hypotheses:
$$ \begin{align} H_1: & \text{ The person is from the college} \\ H_2: & \text{ The person is not from the college} \end{align} $$
And the event:
$$A: \text{ The answer is correct}$$
$$P(H_1) = \frac34, P(H_2)=\frac14, P(A\mid H_1)=1, P(A\mid H_2)=\frac38$$
The probability for $A$ would be:
$$P(A)=P(H_1)P(A\mid H_1) + P(H_2)P(A\mid H_2) = \frac34 + \frac14\cdot\frac38 = 0.84375$$
However, when it comes to part b), I am stuck. How do I proceed?
This can be approached by conditional probability. Let us first compute the probability that a person will give the same answer twice. Notice that either both answers can be wrong or right - as long as they are the same, it counts.
First, the probability that both answers are wrong. This is simple: $0 + \frac{1}{4} * \left(\frac{5}{8}\right)^2 = \frac{25}{256}.$
Next, the probability that both answers are right. This is simply $\frac{3}{4} * 1^2 + \frac{1}{4} * \left(\frac{3}{8}\right)^2 = \frac{201}{256}.$
The requested probability is then $$P = \frac{\frac{201}{256}}{\frac{201}{256} + \frac{25}{256}} = \frac{201}{201 + 25} = \boxed{\frac{201}{226}}.$$