1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.