In essence
probability is influenced by a person's information about the possible occurrence of an event. For example, let the event A be 'I have a new phone'; event B be 'I have a new watch'; and event C be 'I am happy'; and suppose that having either a new phone or a new watch increases the probability of my being happy. Let us assume that the event C has occurred – meaning 'I am happy'. Now if another person sees my new watch, he/she will reason that my likelihood of being happy was increased by my new watch, so there is less need to attribute my happiness to a new phone. To make the example more numerically specific, suppose that there are four possible states \Omega = \left\{ s_1, s_2, s_3, s_4 \right\}, given in the middle four columns of the following table, in which the occurrence of event A is signified by a 1 in row A and its non-occurrence is signified by a 0, and likewise for B and C. That is, A = \left\{ s_2, s_4 \right\}, B = \left\{ s_3, s_4 \right\}, and C = \left\{ s_2, s_3, s_4 \right\}. The probability of s_i is 1/4 for every i. and so In this example, C occurs
if and only if at least one of A, B occurs. Unconditionally (that is, without reference to C), A and B are
independent of each other because \operatorname{P}(A)—the sum of the probabilities associated with a 1 in row A—is \tfrac{1}{2}, while \operatorname{P}(A\mid B) = \operatorname{P}(A \text{ and } B) / \operatorname{P}(B) = \tfrac{1/4}{1/2} = \tfrac{1}{2} = \operatorname{P}(A). But conditional on C having occurred (the last three columns in the table), we have \operatorname{P}(A \mid C) = \operatorname{P}(A \text{ and } C) / \operatorname{P}(C) = \tfrac{1/2}{3/4} = \tfrac{2}{3} while \operatorname{P}(A \mid C \text{ and } B) = \operatorname{P}(A \text{ and } C \text{ and } B) / \operatorname{P}(C \text{ and } B) = \tfrac{1/4}{1/2} = \tfrac{1}{2} Since in the presence of C the probability of A is affected by the presence or absence of B, A and B are mutually dependent conditional on C. == See also ==