"In a multiple-choice test with m options, a student knows the correct answer with a probability p, and in the absence of knowledge, chooses randomly one of the available options. What is the probability that the student knows the correct answer, given that they have answered the question correctly?" Let's clarify things. So, we have k tickets, in each of them there is m possible answers, one of which is correct. Provided the student knows the correct answer to kp of the tickets. He can give a correct answer in two cases:
- He got a good ticket. The probability of this is P1=kp/k=p
- He got a bad ticket, but he hit it. P2=(1-p).1/m 3)By the conditional probability formula, the probability that he knew the answer is P= P1/P1+P2 or P= pm/pm-p+1
I'm getting this far and I don't know if this is the right solution can anyone help with the solution?
Do not make things even more complex. Just think about the question itself.
Let the case "the student answer correctly" be $A$; the case "the student knows the answer" be $B$. We have $$P(B)=p$$ $$P(A|\overline B)=\frac1m$$ $$P(A|B)=1$$ Using conditional probability, we have $$P(A)=P(B)P(A|B)+(1-P(B))P(A|\overline B)=p+\frac{1-p}m$$ Thus, using Bayes' formula $$P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{p}{p+\frac{1-p}m}=\frac{mp}{mp-p+1}$$ Thus, the probability that the student knows the correct answer, given that they have answered the question correctly, is $\frac{mp}{mp-p+1}$