Suppose that Veronica is an economist who is trying to confirm whether or not her theory is true. She has an experimental test that she can run up to $5$ times. Each run of the test either confirms or refutes her theory. If her theory is true, the test confirms it with $80$% probability, and if her theory is false, the test refutes it with $80$% probability. If Veronica becomes convinced that her theory is true, then she stops testing it. In reality, the tests of the theory are independent of each other.
Initially, she is unsure whether her theory is true or not and her prior belief that it is true is $0.5$.
If she runs the test twice and her theory is confirmed both times, what does Bayes rule say her updated belief that the theory is correct should be
I am not really sure about what the question is asking, what are we trying to find out in terms of maths notation?
I could find out
$$P(C|T)=0.8,P(R|F)=0.8\implies P(R|T)=0.2,P(C|F)=0.2\\P(C\cap T)=0.8\times0.5=0.4,\:P(C\cap F)=0.1\implies P(C)=P(R)=0.5$$
You want $P(T \mid C_1 \cap C_2)$, where $C_1, C_2$ are confirmations from tests 1 and 2.
The tests are conditionally independent given whether the theory is true or not, so $$P(C_1 \cap C_2 \mid T) = P(C_1 \mid T) P(C_2 \mid T) = 0.8^2 = 0.64.$$
We also have $$P(C_1 \cap C_2 \mid F) = P(C_1 \mid F) P(C_2 \mid F) = 0.2^2 = 0.04.$$
Bayes' Theorem gives $$P(T \mid C_1 \cap C_2) = \frac{P(C_1 \cap C_2 \mid T) P(T)}{P(C_1 \cap C_2 \mid T) P(T) + P(C_1 \cap C_2 \mid F) P(F)},$$ and working this out we get $$P(T \mid C_1 \cap C_2) = \frac{0.64}{0.64 + 0.04} = \frac{0.64}{0.68} \approx 0.94.$$
Note that "the tests of the theory are independent of each other" is not quite correct - the tests are conditionally independent of each other given whether the theory is true. Once you've seen 4 confirmations in a row, the next test is more likely than not to be a confirmation, so they cannot be unconditionally independent.