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Suppose a student sits for an exam and in a question, there are 5 choices where one is correct. There is also a 1/4 chance that the student may have seen the answer to the question previously. If the student randomly selects an answer for the question if he hadn't seen the answer and selects the correct one if he had seen the answer, what is the probability that the student selects the correct answer? What I think is that this is a conditional probability whereby. the probability of selecting the correct answer is the probability of selecting the correct answer given he had seen the answer or probability of selecting the correct one given he hadn't seen the answer. However, adding the two probabilities gives me a value greater than one. This is because, to me, the probability that he selects the correct answer given he had seen the answer is 1. In what other way can I approach this question?

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    $\begingroup$ en.wikipedia.org/wiki/Law_of_total_probability $\endgroup$ – Mark L. Stone Jul 29 '16 at 14:21
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    $\begingroup$ You are on the right track when you think you need to add the probabilities from the two configurations, but you have the wrong probability on one of the branches of your scenario, fixing this branch's probability should give you the right answer. $\endgroup$ – Antoine Vernet Jul 29 '16 at 14:29
  • $\begingroup$ Thank you Antoine. What branch might be wrong? Am trying to figure it out and am getting back to what I have already $\endgroup$ – Ben Jul 29 '16 at 14:31
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    $\begingroup$ Do yo understand the Law of Total Probability? If not, learn it, and use it. $\endgroup$ – Mark L. Stone Jul 29 '16 at 15:18
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    $\begingroup$ Ben, the example in the link given by @Mark describe a situation that is almost identical to yours. $\endgroup$ – Antoine Vernet Jul 29 '16 at 15:31
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For any given question, you can split it into two situations: the student has seen the question and the student has not.

\begin{align*} P(\mbox{correct})&=P(\mbox{correct}|\mbox{seen})P(\mbox{seen})+P(\mbox{correct}|\mbox{not seen})P(\mbox{not seen})\\ &=1\cdot (1/4)+(1/5)(3/4)\\ &=2/5. \end{align*}

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