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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 the missing probability P(A)0.35 P(AorB)0.55 P(B)?
To find the missing probability ( P(B) ), you can use the formula for the probability of the union of two events: ( P(A \cup B) = P(A) + P(B) - P(A \cap B) ). Assuming ( P(A \cap B) = 0 ) (which is common unless otherwise specified), you can rearrange the equation: ( P(B) = P(A \cup B) - P(A) = 0.55 - 0.35 = 0.20 ). Thus, ( P(B) = 0.20 ).
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).
What element is element B?
Chemical symbol B represents the element Boron. Boron is a p block element.
What is difference between power set and universal set?
From rule of set difference: A \ B = {x is element of A and not element of B} This is a little of first part of the question. When we have set A, set B and finding the difference of P(A) \ P(B) or the same as P(A) - P(B). First we have to make these two power sets of A, and of B. P(A) = { {}, subset of A, other subsets of A, , , (A its self)} P(B) = { {}, subset of B, other subsets of B, , , (B its self)} These two power sets will contain what ever subsets of A, or subsets of B, but first of their elements will be {}, which will be the same. From rule of set difference, I've seen many sample shown P(A) \ P(B) = { {}, subset of A, which not subset of B, , , } The big wonder is {}, the empty set still contained in the result set P(A) \ P(B), even though {} is contained in P(B). It did not being get rid off and other elements if they contained in P(B). Many internets show the same but never explain.
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
Which element contains the first p electron?
The element that contains the first p electron is boron, which has an atomic number of 5. Boron's electron configuration is 1s2 2s2 2p1, meaning that the first p electron is found in the 2p orbital.
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)
