Q. Four roads start from a junction. Only one of them leads to a mall. The remaining roads ultimately lead back to the starting point. A person not familiar with these roads wants to try the different roads one by one to reach the mall. What is the probability that his second attempt will be correct?
My Answer: Let X be the random variable where his second attempt will be correct. So $X\sim BIN(n=4, p=1/4)$ . Does this makes sense? If so, then I'll have
$P(X=2) = f(2) = \left(\begin{array}{c}4\\ 2\end{array}\right)\ (\frac{1}{4})^2(\frac{3}{4})^2 = 27/128$
Am I doing it right?
So, I will explain a solution that does not introduce a binomial random variable. Concretely, there is only one road that will lead him to the mall and three other roads that will not lead him to the mall, but back to the starting position.
The problem is asking: what is the probability $A$ that in his first attempt, he chooses one of these three incorrect roads and in his second attempt he chooses the one correct road.
So, the probability that he chooses an incorrect road is $\dfrac{3}{4}$ and the probability that he chooses a correct road after choosing an incorrect road is $\dfrac{1}{3}$.
So, $P(A)=\dfrac{3}{4}\cdot\dfrac{1}{3}=\dfrac{1}{4}$
Imagine if we wanted to extend this question to the case when he forgets about the road that he chooses. Suppose that we want to find the probability that he finds the correct road in his third attempt. Then, $P(A)=\left(\dfrac{3}{4}\right)^2\cdot\dfrac{1}{4}$ because he has to fail twice before choosing the correct road.
What are these probabilities resembling? Well, since, in this case, if he were to choose the correct road in the $n$th attempt, then $P(A)=\left(\dfrac{3}{4}\right)^{n-1}\cdot\dfrac{1}{4}$. This probability would resemble a geometric distribution because it involves him failing a certain number of times before succeeding.