Stats Problem via Probability of Guessing on a test There are 5 multiple choice questions on an exam, each with 5 possible answers, whats the probability of guessing and getting all 5 right? And what is the probability of getting all 5 wrong...
Im less interested in the answer, and more interested on how to solve this problem
 A: Assuming your probability of getting each problem correct is independent of the others, then the probability of getting them all correct is $\frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}$.
Similarly the probability of getting them all incorrect would be $\frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}$
A: There are two ways that I think about it, a graphical way and a direct computation.  I'll tell you a graphical way first, because I think it helps to understand the second way.
Decision tree
I'd think of it as a decision tree.  For the first question, you have five choices, and you have to choose one.  So you choose each with $p=\frac{1}{5}$.  One of those is the right choice, four of them are wrong choices.  So draw a little tree in your head, with the five answers as vertices attached to the root, and $p=\frac{1}{5}$ on each edge.
The second question is independent of the first question, so you can treat each of the five vertices from the first question as the root of a subtree for the second question.  Each of the five subtrees again has a root and five vertices, each of which are attached to the root with an edge labeled with $p=\frac{1}{5}$.  And so on for the remaining questions, yielding a tree with five levels.
Each of the leaves of this tree identifies a path through the decision tree back up to the root, which is the same as a series of answers.  There are $5^5=3125$ leaves.  
Only one of them is associated with a series of five correct answers, since each question has one correct answer, so the probability of successfully guessing all five answers is $\frac{1}{3125}$.  We could do a similar exercise to count the cases where all answers are incorrect, but it's not easy to see.  If that number is $x$, then the probability of choosing all wrong answers is $\frac{x}{3125}$.
Direct computation
If we have independent events, the probability of a union of those events occuring is the product of their probabilities. So if we're choosing n answers, each of which has m choices, the probability that we choose the first one correctly is $p=\frac{1}{m}$, and the probability that we choose each correctly is $p=\frac{1}{m}^n$.  The probability that we choose one incorrectly is $p=\frac{m-1}{m}$, and each incorrectly is $p=\frac{m-1}{m}^n$
