Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The pupil knows how to translate the word "art" from English to Belarusian. That's true, teacher knows it. In a year, the teacher wants to check if the pupil still remember the translation of the word. So, the teacher gives him a multiple choice quiz to answer this question. There are 5 options to answer, only one is correct. The correct answer was A and the pupil decided to choose A as an answer (anyway he should choose any answer among 5 options).

So, what is the probability the pupil REALLY remembers how to translate "memory"?

My guessing... Let me started, please.

Lets say the event the pupil remembers how to translate the word is R. We have to find P(R) - the probability the pupil still remembers the translation.

Let's A is the event the person chooses A in the quiz.

We have two conditional probabilities:

P (A|not R)=0.2 (he doesn't know the answer)

P (A|R)=1 (if he knows how to translate, he answers A, because that is the correct answer).

We have to find P(R).

Can we or we can't?

I think we can actually add P(A)=P(B)=P(C)=P(D)=P(E)=0.2, if that helps (any of 5 options A,B,C,D,E could be correct, there is no more information about probable answer).

share|cite|improve this question
If you're willing to specify your prior belief about the probability that the student remembers, then you can use Bayes' Formula to update and obtain the posterior probability that he remembers given that he answered correctly. Otherwise, as kaine notes, you cannot reach any conclusion. – Jonathan Christensen Dec 20 '12 at 22:09
up vote 3 down vote accepted

If the only alternatives are $R$ (he remembers the right word) and $R^c$ (he forgets the word completely and just takes a random guess), then you can say that, if $A$ is the event of getting the correct answer, $$P(A) = P(A|R) P(R) + P(A|R^c) P(R^c) = P(R) + \frac{1}{5} (1 - P(R)) = \frac{1}{5} + \frac{4}{5} P(R) $$ so that $$P(R) = \frac{5 P(A) - 1}{4}$$ Thus if you gave a test with this type of question for $n$ different words, and the student got $x$ answers correct, an unbiased estimator for the fraction of those $n$ words that the student remembered would be $(5 x/n - 1)/4$.

The trouble is, remembering is not an all-or-nothing matter. Perhaps the student wouldn't remember мастацтва (that's what Google Translate gives as Belarusian for art) if simply asked to write the answer, but seeing it as one of the alternatives jogs his memory, or he knows words (b) and (d) mean something else.

share|cite|improve this answer

No we cannot.

There are too many unknowns to make any such assertion. All we can conclusively say is that he answered the question correctly on the test. I believe it is possible to estimate average retention, however, if you have several students under the same conditions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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