Compound Events Basics of Probability 1 Rules Axioms

Compound Events Basics of Probability 1

Rules (Axioms) of Probability � Axioms Of probability 1. P(S) = 1 2. 0 ≤ P(E) ≤ 1 3. For each two events E 1 and E 2 with E 1 ∩ E 2 = Ø, P(E 1 U E 2) = P(E 1) + P(E 2) (disjoint addition rule) � 1&2 imply: P(Ø) =0 and P(E’) = 1 – P(E) (Compliment Rule) 2

Compound events � The Union of two events is the event consisting of all outcomes that are contained in either of two events, denoted: E 1 ⋃ E 2. Called E 1 or E 2. � The Intersection of two events is the event consisting of all outcomes that contained in both of two events, E 1 ⋂ E 2. Called E 1 and E 2

Mutually Exclusive Events � Two events A and B are mutually exclusive (disjoint) if they have no outcomes in common. Therefore they cannot occur simultaneously. � If events A and B are mutually exclusive: (Axiom 3) ◦ Symbolically, A ∩ B = Ø ◦ Then P=(A U B) is simply: P(A) + P(B) 4

Non Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) When finding the Union of two events, we need to be careful not to count shared outcomes (the intersection) more than once.

Finding A Union � The Addition Rule: P(A∪B) = P(B)+P(A) – P(A∩B) � If events A and B are mutually exclusive: (Axiom 3: Disjoint addition Rule)

Finding An Intersection � We know if A and B are disjoint then: P(A∩B)= 0 � We could rewrite the Addition Rule: P(A∩B) = P(B)+P(A) – P(A∪B) � We more often use the multiplication rule. P(A ∩ B) = P(B|A)*P(A) = P(A|B)*P(B) � The last expression is obtained by exchanging the roles of A and B. 7

Finding Conditional Probabilities � 8

Independence and Conditionals � If we have two independent events A and B then: � P(A|B) � For = P(A) and P(B|A) = P(B) Example let: ◦ P(O) be the probability you oversleep ◦ P(B) be the probability you eat breakfast. ◦ P(G) be the probability you have green eyes � P(B|O) is of interest. � P(B|G) is simply the same as P(B). 9

Independent Multiplication Rule � If two events A and B are independent, then: P(A|B)=P(A) � So: P(A ∩ B) = P(A|B)*P(B) � Becomes: P(A ∩ B) = P(A) x P(B) 10

Relationships and Rules � Recall of our possible relationships of two events: Independent? Yes No Yes 1 2 No 3 4 Disjoint? � We need to know how to find Unions and intersections in all 4 of these regions 11

Generalizing Unions � We know how to find P(A ⋃ B) if the are Disjoint ◦ What if we want to know P(A ⋃ B ⋃ C)? �P(A)+P(B)+P(C) ◦ Extended to P(A ⋃ B ⋃ C ⋃ D…. . ⋃ N )? �P(A)+P(B)+P(C)+P(D)…. +P(N) � We also know how to find P(A ⋃ B) if the are Not Disjoint ◦ What if we want to know P(A ⋃ B ⋃ C)? �P(A)+P(B)+P(C)- P(A ⋂ B)- P(A ⋂ C)-P(B ⋂ C) + P(A ⋂ B ⋂ C) ◦ Extended to P(A ⋃ B ⋃ C ⋃ …. . )? �Messy +/- intersections

Generalizing Intersections � We also know how to find P(A ⋂ B) if they are Independent ◦ What about P(A ⋂ B ⋂ C)? �P(A)*P(B)*P(C) ◦ Extended to P(A ⋂ B ⋂ C ⋂ D…. . ⋂ N )? �P(A)*P(B)*P(C)*P(D)…. . * P(N) � We also know how to find P(A ⋂ B) if the are Dependent ◦ What about P(A ⋂ B ⋂ C)? ◦ Extended to P(A ⋂ B ⋂ C ⋂ D…. . ⋂ N )? �Have to deal with multiple conditionals 13

Addition Rule Example A Company’s staff is organized in the following table: Staff Females Males Total Salary 7 1 8 Hourly 3 2 5 Total 10 3 13 If a staff person is randomly selected, find the probability that the subject is on salary or a male. P(Salary) + P(Male) – P(Salary and Male)

Addition Rule Example � Pick a card from a deck, what is the probability that the card is a 4 or a king? � Are these events mutually exclusive? ◦ Yes � Event 4 = picking a 4 P(4) = 4/52 � Event K = picking a king P(K) = 4/52 � P(4 or K) = P(4) + P(K) = 4/52 + 4/52 = 2/13 � What if these events were not mutually exclusive? ◦ Find P(K or Heart):

Finding a Conditional Probability � 16

Multiplication Rule Example � 17

Independent Multiplication Rule Example �A coin is tossed and a die is thrown, find the probability of getting a tail on the coin and a 3 on the die. � Are these events independent? ◦ Yes (a coin toss will not affect a die throw) ◦ Event T is getting a tail = P(T) = 1/2 ◦ Event 3 is getting a 3 on the die. =P(3) = 1/6 � So: P(T and 3) = P(T)P(3)=(1/2)(1/6) = 1/12 18

Multiple dependent events A production lot of 850 parts contains 50 defectives. Two parts are selected at random (w/o replacement). � Now, 3 parts are sampled randomly. � What is the probability of drawing this sequence: the first two are defective, while third is not? 19

Multiple independent events � The probability that a running back fumbles the ball on any given play is 0. 01. Assume each time he carries the ball is independent. � If the running back has 15 carries in a game, what is the probability that he DOES NOT fumble? P(E 1 ∩ E 2 ∩ … ∩ Ek) = P(E 1)*P(E 2)*…*P(Ek)= (0. 99)15 = 0. 86. 20