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Formulas on Percentage Base and Rate?
P=B×RB=P÷RR=P÷B
How might two probabilities be added?
If A and B are two events then P(A or B) = P(A) + P(B) - P(A and B)
What is the Formula for odds for either of two events?
If A and B are mutually exclusive event then Probability of A or B is P(A)+P(B). If they are not mutually exclusive then it is that minus the probability of the P(A)+P(B) That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it is clear that if they are mutually exclusive, P(A and B)=0 and we have the first formula.
What is addition theorem of probability?
Consider events A and B. P(A or B)= P(A) + P(B) - P(A and B) The rule refers to the probability that A can happen, or B can happen, or both can happen together. That is what is stated in the addition rule. Often P(A and B ) is zero, if they are mutually exclusive. In this case the rule just becomes P(A or B)= P(A) + P(B).
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
What are the release dates for B-P- Birthday Present - 2004?
B-P- Birthday Present - 2004 was released on: USA: 15 October 2004 (Script-2-Screen Film Festival)
What the notes for happy birthday on steel pan?
h a p p y b I r t h d a y
If A and B are independent events then are A and B' independent?
if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof
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.
What is the formula for inclusive events?
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.
How do you find P A given B?
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)
Addition rule for probability of events A and B?
If they're disjoint events: P(A and B) = P(A) + P(B) Generally: P(A and B) = P(A) + P(B) - P(A|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).
Definition of additive law in probability?
This has to do with the union of events. If events A and B are in the set S, then the union of A and B is the set of outcomes in A or B. This means that either event A or event B, or both, can occur. P(A or B) = P(A) + P(B) - P(A and B) **P(A and B) is subtracted, since by taking P(A) + P(B), their intersection, P(A and B), has already been included. In other words, if you did not subtract it, you would be including their intersection twice. Draw a Venn Diagram to visualize. If A and B can only happen separately, i.e., they are independent events and thus P(A and B) = 0, then, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - 0 = P(A) + P(B)
Formulas on Percentage Base and Rate?
P=B×RB=P÷RR=P÷B
How might two probabilities be added?
If A and B are two events then P(A or B) = P(A) + P(B) - P(A and B)
What is the Formula for odds for either of two events?
If A and B are mutually exclusive event then Probability of A or B is P(A)+P(B). If they are not mutually exclusive then it is that minus the probability of the P(A)+P(B) That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it is clear that if they are mutually exclusive, P(A and B)=0 and we have the first formula.
