A statement with probability $x$ of being true is given. Two independent people (A and B) are brought in to assess the validity of said statement. The probability that A is correct is $p_1$. The probability that B is correct is $p_2$. They both agree that the statement is true. What is the probability that the statement is, in fact, true?
$\endgroup$ 32 Answers
$\begingroup$What we want to calculate is the conditional probability of the statement beeing true, given A and B saying it is true.$$ P(\text{the statement is correct}| \text{A and B say it is true}) = \frac{P(\text{A and B say it is true} | \text{the statement is correct})P(\text{the statement is correct})}{P(\text{A and B say it is true})} = \frac{P(\text{A and B say it is true} | \text{the statement is correct})P(\text{the statement is correct})}{P(\text{A and B say it is true}| \text{the statement is correct})+ P(\text{A and B say it is true}| \text{the statement is not correct})} $$Hier is the problem, I don’t see a way to access $P(\text{A and B say it is true}| \text{the statement is not correct})$.
$\endgroup$ 4 $\begingroup$I belive I may have cracked it.
Let $A_T$: "A says it's true"
$\ \ \ \ \ $ $B_T$: "B says it's true"
$\ \ \ \ \ $ $X_T$: "The statement is true"
$P(X_T|A_T \cap B_T) = \frac{P(A_T \cap B_T|X_T) \times P(X_T)}{P(A_T \cap B_T)} = \frac{P(A_T \cap B_T|X_T) \times P(X_T)}{P(A_T \cap B_T \cap X_T) + P(A_T \cap B_T \cap \overline{X_T})}$
$P(A_T \cap B_T|X_T)^{*_1} = p_1p_2$
$P(A_T \cap B_T \cap X_T) = P(A_T \cap B_T|X_T) \times P(X_T) = p_1p_2x$
$P(A_T \cap B_T \cap \overline{X_T}) = P(A_T \cap B_T|\overline{X_T})^{*_2} \times P(\overline{X_T}) = (1-p_1)(1-p_2)(1-x)$
Finally,
$$
P(X_T|A_T \cap B_T) = \frac{p_1p_2x}{p_1p_2x+(1-p_1)(1-p_2)(1-x)}
$$
*1: The probablity of A and B being right
*2: The probablity of A and B being wrong
