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.

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    $\begingroup$ Can you work out the probability that the student gets no questions correct? Can you see how you then find the probability you want? $\endgroup$ – David Jul 22 '14 at 4:57
  • $\begingroup$ Assuming the student guesses? $\endgroup$ – Thomas Andrews Jul 22 '14 at 5:04
  • $\begingroup$ @ThomasAndrews Yes $\endgroup$ – Ovi Jul 22 '14 at 5:05
  • $\begingroup$ @David It wouldn't be $(\dfrac 45)^4$, would it? $\endgroup$ – Ovi Jul 22 '14 at 5:08
  • $\begingroup$ That's the answer to my first question, now can you find the answer you need? What is the relation between the two questions? $\endgroup$ – David Jul 22 '14 at 5:10

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 $$P(\textrm{at least 1 answer right)}=1-P(\textrm{no answers right})=1-\left(\frac{4}{5}\right)^{4}$$


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