Given three events \(A\), \(B\), and \(C\), we say that \(A\) and \(B\) are conditionally independent given \(C\) if the occurrence of one event does not affect the probability of the other event occurring, given that \(C\) has occurred. This can be expressed mathematically as:

\[P(A \cap B \given C) = P(A \given C) P(B \given C)\]

The formula can be rearranged to a more intuitive form. First, observe that the LHS can be rewritten as,

\[\begin{align} \begin{aligned} P(A \cap B \given C) & = \frac{P(A \cap B\cap C)}{P(C)}\\ & = \frac{P(A\cap C)P(B\given A\cap C)}{P(C)} \end{aligned} \end{align}\]

Now transform the RHS as,

\[\begin{align} \begin{aligned} P(A \given C) P(B \given C) & = \frac{P(A\cap C)}{P(C)} P(B\given C) \end{aligned} \end{align}\]

Re-drawing the equality with the transformed LHS and RHS, we get,

\[\begin{align} \begin{aligned} \frac{P(A\cap C)P(B\given A\cap C)}{P(C)} = \frac{P(A\cap C)}{P(C)} P(B\given C) \end{aligned} \end{align}\]

which, after cancellations, gives,

\(\begin{align} \begin{aligned} P(B\given A\cap C)= P(B\given C) \end{aligned} \end{align}\) which more explicitly states that “when you know that \(C\) has occurred, learning that \(A\) has occurred does not provide any information on the probability of \(B\) occurring”.