I want to solve the following exercise:
John tells the truth two out of every three times he is asked, while Peter tells the truth four out of every five times. Both agree in assuring that from a box of marbles containing 6 marbles of different colors a red one was taken. Determine the probability that this statement is true.
So, consider: A:John tells the truth
B:Peter tells the truth
C:A red marble comes out of the marble box.
I think our objective is to determine
$$P(E \mid A \cap B)=\frac{P(A \cap C \mid E) P(E)}{P(A \cap C)}$$
Thinking that $P(A)= \frac{2}{3}$, $P(B)= \frac{4}{5}$
And using the total probability theorem I could determine $P(A \cap C)$
I really don't see clearly how to perform the exercise approach, it is a bit confusing. Any help?
The events you should be looking at are:
$J$: John says the marble was red
$P$: Peter says the marble was red
$R$: The marble was red
Then the probability we are trying to calculate is $P(R | J \cap P)$, i.e. the probability the marble is actually red given that John and Peter both say that it was red.
From Bayes' theorem, you can express this as:
$$P(R | J \cap P) = \frac{P(J \cap P | R) P(R)}{P(J \cap P)}$$
$P(R)$ should be fairly simple to calculate, while $P(J \cap P)$ will involve some decomposition into parts of $P(J \cap P \cap R)$ and $P(J \cap P \cap R^c)$.