Q: A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4
choices seem equally plausible.
If they know the answer they will get the question
right. If not they have to guess from the 3 or 4 choices.
As the teacher you want the test to measure what the student knows. If the student
answers a question correctly what’s the probability they knew the answer?
A: Stuck in the process of trying to answer this.
The way I thought about it was using Baye's rule and treating it as a conditional probability.
In that, $$P(answer correct | knew answer) = \frac{P(know answer | answer correct) \cdot P(answer correct)}{P(knew answer)}$$
The probability that the student knows the answer is 0.5 so that gives us the denominator. Figuring out probability for the answer being correct and knowing the answer given a correct answer is a bit more confusing for me. Would appreciate guidance.
Best Answer
With
$A$: students answers correctly
$B$: student knew answer
what you want is $P(B|A)$, rather than $P(A|B)$
Now, we are given that:
$$P(B)= \frac{1}{2}$$
and hence
$$P(B^C)=1-\frac{1}{2}=\frac{1}{2}$$
Also:
$P(A|B)=1$ (if they knew the answer they give the correct answer)
and
$$P(A|B^C)=\frac{1}{4}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{4}=\frac{1}{12}+\frac{3}{16}=\frac{4}{48}+\frac{9}{48}=\frac{13}{48}$$
(When they don't know the answer there is a $\frac{1}{4}$ probability they can eliminate one of the answers and thus have a $\frac{1}{3} $ probability of guessing correctly between the remaining 3, and there is a $\frac{3}{4}$ probability they can't eliminate any one answer in which case they have a $\frac{1}{4}$ probability of guessing correctly)
Next, we have that:
$$P(A \cap B)=P(A|B)\cdot P(B)=1\cdot \frac{1}{2}=\frac{1}{2}$$
and
$$P(A \cap B^C)=P(A|B^C)\cdot P(B^C)=\frac{13}{48}\cdot \frac{1}{2}=\frac{13}{96}$$
and thus:
$$P(A)=P(A \cap B)+P(A \cap B^C)=\frac{1}{2}+\frac{13}{96}=\frac{61}{96}$$
So finally:
$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}}{\frac{61}{96}}=\frac{48}{61}$$