P(A|B)= P(A n B) / P(B)
P(A n B) = probability of both A and B happening
to check for independence you see if P(A|B) = P(B)
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P(A given B')=[P(A)-P(AnB)]/[1-P(B)].
P(A given B')=[P(A)-P(AnB)]/[1-P(B)].In words: Probability of A given B compliment is equal to the Probability of A minus the Probability of A intersect B, divided by 1 minus the probability of B.
Given two events, A and B, the conditional probability rule states that P(A and B) = P(A given that B has occurred)*P(B) If A and B are independent, then the occurrence (or not) of B makes no difference to the probability of A happening. So that P(A given that B has occurred) = P(A) and therefore, you get P(A and B) = P(A)*P(B)
P(A'/B)=P(A'nB)/P(B)
Let's try this example (best conceived of as a squared 2x2 table with sums to the side). The comma here is an AND logical operator. P(A, B) = 0.1 P(A, non-B) = 0.4 P(non-A, B) = 0.3 P(non-A, non-B) = 0.2 then P(A) and P(B) are obtained by summing on the different sides of the table: P(A) = P(A, B) + P(A, non-B) = 0.1 + 0.4 = 0.5 P(B) = P(A,B) + P(non-A, B) = 0.1 + 0.3 = 0.4 so P(A given B) = P (A, B) / P (B) = 0.1 / 0.4 = 0.25 also written P(A|B) P(B given A) = P (A,B) / P (A) = 0.1 / 0.5 = 0.2 The difference comes from the different negated events added to form the whole P(A) and P(B). If P(A, non-B) = P (B, non-A) then P(A) = P(B) and also P(A|B) = P(B|A).