What is the probability that he really knows the answer

[zorro]

Question:
In answering a question on a multiple choice test , an examine either knows the answer or he guesses or he copies. Suppose each question has 4 choices. Let the probability that examine copies the answer is 1/6 and the probability that he guesses is 1/3. The probability that his answer is correct given that he copied the answer is 1/8. Suppose an examine answers a question correctly, what is the probability that he really knows the answer

What approch should i take to start solving this problem . Should i use the Bayes theorem or no? please provide with the name of the procedure / theorem

[jk22]

Hello, nice to meet you. I'm new in this forum.

Yes, this is Bayes thm, the steps made be like : p(know)+p(copy)+p(guess)=1

p(knows)=1-1/3-1/6=1/2

then p(knows|correct)=p(correct|knows)*p(knows)/p(correct)
=1*1/2*1/(p(correct|knows)p(knows)+p(correct|guess)p(guess)+p(correct|copy)p(copy))
=.5*1/(p(knows)+p(guess)+1/48)
=24/41

If we read the last sentence, after the coma only, it depends ? if he knows the answer, the answer is forcedly correct, as well as guess.

maybe this could help.

c ya.

[Martingale]

then p(knows|correct)=p(correct|knows)*p(knows)/p(correct)
=1*1/2*1/(p(correct|knows)p(knows)+p(correct|guess)p(guess)+p(correct|copy)p(copy))
=.5*1/(p(knows)+p(guess)+1/48)
=24/41

That number seems a little low.

[zorro]

then p(knows|correct)=p(correct|knows)*p(knows)/p(correct)
=1*1/2*1/(p(correct|knows)p(knows)+p(correct|guess)p(guess)+p(correct|copy)p(copy))
=.5*1/(p(knows)+p(guess)+1/48)
=24/41

That number seems a little low.

If u know the answer can u please let me know ?

[jk22]

Hello, nice to meet you.

you mean numerically, or do you mean if he answered correctly, we could think he knew really ? (1 seems here a bit naive/gentle seen the other data), except if we define "correctly" as neither copied nor guessed.

(the probability it was just guessed, (maybe it can be discussed on the meaning of this word, either : taken as : put at random, like after dice throwing, the correct one, without knowing, or with some intuition/instinct evident from itself, which would then be taken as equivalent to "knowing"), is 1/3, dumbly copied (hence without knowing), 1/6*1/8, which leaves
31/48 at most, among them, without random nor copying some people answered not correctly)

remain the data : 4 which was not used.