Why aren't these two probability statements the same? If each question has only one answer, and four options; and you want to get at least three out of five answers correct then there are two answers I can think of, both using choose as the binomial function.
Option A
$$\binom{5}{3} * (\frac{1}{4})^3$$
Option seems to guarantee that you will get at least three questions right, with a probability of $15.6$%
Option B
Here we split it into case statements:
When we want three questions correct:

$$\binom{5}{3} * (\frac{1}{4})^3 * (\frac{3}{4})^2$$
When we want four questions correct:

$$\binom{5}{4} * (\frac{1}{4})^4 * (\frac{3}{4})$$
When we want five questions correct:

$$(\frac{1}{4})^5$$
The only problem is when we add up all of the above cases, we get around $10$% as the probability.
So my question is, which method is wrong? Which method is right? How come Option A does not give the same solution as Option B? Why is it different? To me, they should both give the same solution?
 A: The problem lies in Situation A:  $(\frac{1}{4})^3$ does represent picking the correct answer at random for three different problems. The $5\choose3$ does represent the different ways to order those three correct choices among the five problems.  You are not adressing the last two incorrect (or correct) answers.  Leaving them off is almost like including a $(1)^2$ at the end, which would mean that the probability of the other two questions being answered wrong is 100%- which is also why your probability is larger.  Situation B is accurate because it addresses the other problems that may or may not be correct.
A: The second is correct, and the first is not. We will try to explain why the answer (1) overcounts.
Let the questions be numbered 1, 2, 3, 4, and 5. The idea behind the first formula seems to be as follows. The probability of getting 1, 2, and 3 (and possibly others) right is $\left(\frac{1}{4}\right)^3$. The probability of getting 1, 2, and 4 (and possibly others) right is also $\left(\frac{1}{4}\right)^3$.  The same is true for every one of the $\binom{5}{3}$ choices of $3$ questions from 1, 2, 3, 4, 5. So in effect you added together $\binom{5}{3}$ copies of $\left(\frac{1}{4}\right)^3$.
However, consider the $4$ questions 1, 2, 3, 4. In adding up, we counted repeatedly the situations where we get all these right. It was counted in our expression for the probability we get 1, 2, 3, and possibly others. It was also counted in our expression for the probability that we get 1, 2, 4, and possibly others, right. It was also counted in our expression that we get at least 1, 3, 4, and possibly others. And it was counted in our expression for the probability that we get at least 2, 3, and 4 right. 
With care, one can compensate for the overcounting. That idea is at the heart of the Principle of Inclusion/Exclusion. However, compensating for overcounting is delicate, and the division into disjoint cases is, for this problem, easier. 
A: Option A is wrong. You will still be answering the other questions. 
