I read this question recently: Probability that a student knows the answer
Putting this in reverse, if the student guessed the answer then the probability would be:
\begin{eqnarray*}
A &=& \mbox{Student knows the correct answer} \\
C &=& \mbox{Student answered correctly.} \\
\end{eqnarray*}
GIVEN: Probability that the student knows the answer is 2/3
GIVEN: Probability that the student guesses an answer and is correct is 1/4
Finding $P(A^C \mid C)$ using Bayes' theorem:
\begin{eqnarray*}
P(A^C \mid C) &=& \dfrac{P(C \mid A^C )P(A^C )}{P(C \mid A)P(A) + P(C \mid A^c)P(A^c)} \\
&& \\
&=& \dfrac{\frac{1}{3} \times \frac{1}{4}}{1 \times \frac{2}{3} + \frac{1}{4} \times \frac{1}{3}} \\
&& \\
&=& \dfrac{1}{9}
\end{eqnarray*}
So my question is, if the student did a 2nd or $nth$ test and it had the same question, how would you calculate the probability that the student guessed the answer for both tests assuming he got the answer right on both times?
I'm assuming that the student doesn't know the results (or even remember the previous test) so he's basically just getting lucky doing the same guess and being correct each time he does the test question. So since each test taken is independent of each other, does that mean that everything can essentially be taken to the power of 2 for two tests or to the $nth$ power for $n$ number of tests taken?
\begin{eqnarray*}
D &=& \mbox{Student knows the correct answer for n questions} \\
E &=& \mbox{Student answered correctly for n questions} \\
\end{eqnarray*}
$$P(D)=P(A)^n ?$$
$$P(E)=P(C)^n ?$$
Then you could follow through that the probability that the student is guessing the answer, after we witnessed them being all correct, can be calculated as:
$$P(D^C \mid E)=P(A^C \mid C)^n=(\dfrac{1}{9})^n$$
Right?
EDIT: Ignore $(D)$ and $(E)$ for now (if that is helpful) and do in terms of $(A)$ and $(C)$ for two tests (or more). Here's some helpful visualization:
So the blue circle shows the probability path I'm trying to find out.
Rephrasing a little better: What is the probability that the student is guessing the answer for 2 tests (or $n$ tests) after witnessing the event C (that he answered correctly)?
Best Answer
Unfortunately, it's not that simple. You have to use Bayes' Theorem again. Firstly, since all answers are made independently of each other, given that the student is guessing, the probability of answering all $n$ questions correctly is $P(E \mid D^c) = \left(\frac{1}{4}\right)^n$. So,
\begin{eqnarray*} P(D^c \mid E) &=& \dfrac{P(E \mid D^c)P(D^c)}{P(E \mid D^c)P(D^c) + P(E \mid D)P(D)} \\ && \\ &=& \dfrac{\left(\frac{1}{4}\right)^n \times \frac{1}{3}}{\left(\frac{1}{4}\right)^n \times \frac{1}{3} + 1 \times \frac{2}{3}} \\ && \\ &=& \dfrac{1}{1 + 2\times 4^n}. \end{eqnarray*}