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).
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)
to find the rate divide the percentage by the base that is R=P divide B OR R=P/B THEN CHANGE THE ANSWER TO PERCENT
P(A'/B)=P(A'nB)/P(B)
This is a tricky question, but if you study probability, re-writing it in mathematical notations and using simple definitions and algebra would show you that the answer is yes. We could rewrite the question as: if B occur, A is more likely to occur, Probability of A happening given B happened > probability of A happening P( A given B) > P(A) Is it true that in this case, P(B given A) > P(B)? by definition, P( B given A) = P(A and B) / P( A) = P( A given B) * P(B) / P(A) Since P( A given B) > P(A), then P( A given B)/ P(A) > 1 thus P( B given A) > P(B) and the answer is true, occurence of A makes B more likely. Say A = lung cancer, B = smoking. If someone is smoking, they are more likely to have lung cancer. If someone has lung cancer, they are more likely to be a smoker or have smoked (compared to being a non-smoker).
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).
The probability of inclusive events A or B occurring is given by P(A or B) = P(A) + P(B) - P(A and B), where P(A) and P(B) represent the probabilities of events A and B occurring, respectively.
Sum Rule: P(A) = \sum_{B} P(A,B) Product Rule: P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B) [P(A|B) means probability of A given that B has occurred] P(A, B) = P(A) P(B) , if A and B are independent events.
P(A)=.35 P(B given A)=0.6 P(A and B)= ?
P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).
The rate usually is the percentage.