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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 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)
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)
What does a line reflection not preserve?
A reflection in a line l is a correspondence that pairs each point in the plane and not on the linewith point P' such that l is the perpendicular bisector of segment PP'. IF P is on l then P is paired with itself ... Under a reflection the image is laterally inverted. Thus reflection does NOT preserve orientation...
How do you find the coordinates of a quadrilateral after reflecting it over the y axis?
If the coordinates of a point, before reflection, were (p, q) then after reflection, they will be (-p, q).
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)
Do all figures with rotational symmetry also have reflection symmetry?
No.For example, a hexagon with equal angles and sides of lengths a,b,a,b,a,b has rotational symmetry of order 3, but it has no reflection symmetry.No.For example, a hexagon with equal angles and sides of lengths a,b,a,b,a,b has rotational symmetry of order 3, but it has no reflection symmetry.No.For example, a hexagon with equal angles and sides of lengths a,b,a,b,a,b has rotational symmetry of order 3, but it has no reflection symmetry.No.For example, a hexagon with equal angles and sides of lengths a,b,a,b,a,b has rotational symmetry of order 3, but it has no reflection symmetry.
Formulas on Percentage Base and Rate?
P=B×RB=P÷RR=P÷B
What are the coordinates of the reflection of (a b) over the y-axis?
They are (-a, b).
