A student is taking a multiple choice test where each question has four options for an answer. The student mastered mastered 70% of the material. Assume this means that the student has a 0.7 chance of knowing the correct answer to a random test question. On the other hand, if the student does not know the answer to the question, she randomly selects among the four answer choices. Finally, assume that this holds for each question independent of the others.

What is the probability that a specific question is answered correctly?

P(correct) = 0.7+0.25*0.3 = 0.775

which also means the the P(incorrect) = 1-0.775 = 0.225

I know this is the way to calculate the answer, but I can figure out what the method/formula is used. I was thinking some sort of Binomial distribution but the thing with having an "extra chance" with 0.25*0.3 confuse me.

  • 1
    $\begingroup$ The form mutually exclusive and exhaustive set of events. So $P(\text{knows,correct}) + P(\text{guess,correct}) + P(\text{guess,incorrect}) = 1$ and it is given $P(\text{knows,correct}) = 0.7$. $\endgroup$
    – jonsno
    Commented Aug 20, 2017 at 13:32

1 Answer 1


Let $C$ denote the event that the question is answered correctly and let $K$ denotes the event that the student knows the answer.

Practicized is then that:$$P(C)=P(C\cap K)+P(C\cap K^{\complement})=P(C|K)P(K)+P(C|K^{\complement})P(K^{\complement})=1\cdot0.7+0.25\cdot0.3$$

Here $K^{\complement}$ stands for the event that the student does not know the answer. In that case she randomly chooses out of $4$ so that $P(C|K^{\complement})=0.25$.


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