Probability of getting a right answer? A student is taking a $4$ question multiple choice quiz with each question having $5$ options. What is the probability that he will get at least one question correct?
P.S. Please keep answers at basic statistics level
I'm really not sure what I should do for this problem. I tried various approaches including listing all the possibilities, but then I realized there were many more possibilities than I thought. I also thought of simplifying the question to a quiz of $2$ questions with $2$ choices each, but that didn't really give me any insight.
 A: First thing to notice to make problem a lot easier is to notice that 
$$P(\textrm{at least 1 answer right)}+P(\textrm{no answers right})=1$$
since you either get no answers right or you had to get at least one right with a 100% probability now from this we see that 
$$P(\textrm{at least 1 answer right)}=1-P(\textrm{no answers right})$$
So if you can calculate $P(\textrm{no answers right})$ your answer will come easy. Now in order to calculate $P(\textrm{no answers right})$ notice that each time you answer a question you have $\frac{1}{5}$ probability of getting it right and also answering previous questions right or wrong doesn't change probability of answering next question right or wrong (i.e. independent events). thus 
$$P(\textrm{no answers right})=P\left(\bigcap_{i=1}^{4}(\textrm{Question i is wrong})\right)=\prod_{i=1}^{4}P(\textrm{Question i is wrong})=\left(\frac{4}{5}\right)^{4}$$
If this seems like a little bit of a stretch to you. You could also take a combinatorics approach. This approach is good when each possible outcome is just as equally likely to happen which seems reasonable in this situation so $P(\textrm{no answers right})=\dfrac{\textrm{# of combos of possible answers so get all wrong}}{\textrm{# of totals combos of possible answers}}$
where for "# of combos of possible answer so get all wrong" for each answer there are 4 answers you could write down that could be wrong for each question (since 1 write answer for each question) so # of combos of possible answers so get all wrong=$4^{4}$ (to verify that this equals all those combos think of it in terms of tree diagrams) thus we have from this logic
$$P(\textrm{no answers right})=\dfrac{\textrm{# of combos of possible answers so get all wrong}}{\textrm{# of totals combos of possible answers}}=\frac{4^{4}}{5^{4}}=\left(\frac{4}{5}\right)^{4}$$ 
Thus finally we have 
$$P(\textrm{at least 1 answer right)}=1-P(\textrm{no answers right})=1-\left(\frac{4}{5}\right)^{4}$$
