0
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)
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
How do you find A complement given B complement?
P(A given B')=[P(A)-P(AnB)]/[1-P(B)].
How does one find the probability of A given B compliment?
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.
What is multiplication rule in probability?
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)
Probability of A compliment given by B?
P(A'/B)=P(A'nB)/P(B)
Give the example of why probabilities of A given B and B given A are not same?
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).
How do you find A complement given B complement?
P(A given B')=[P(A)-P(AnB)]/[1-P(B)].
How does one find the probability of A given B compliment?
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.
What is multiplication rule in probability?
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)
How do you find the rate when the base and percentage are given?
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
Probability of A compliment given by B?
P(A'/B)=P(A'nB)/P(B)
If the occurrence of B makes A more likely does the occurrence of A make B more likely?
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).
Give the example of why probabilities of A given B and B given A are not same?
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).
What is the product rule and the sum rule of probability?
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.
How do you solve for the probability of dependent events?
P(A)=.35 P(B given A)=0.6 P(A and B)= ?
How do you find the probability of independent and dependent events?
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...).
How do you get the rate when the base and percentage is given?
The rate usually is the percentage.
How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?
There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12
