BCOR 1020 Business Statistics Lecture 7 – February 7, 2007

Overview Chapter 5 – Probability –Rules of Probability –Conditional Probability –Independent Events

Venn Diagrams Defined… A Venn Diagram is a visual tool for representing sets which will aid our intuition is discussing probabilities and their properties. In a Venn Diagram, the universal set (set of all possible events) is represented as the interior of a rectangle. Subsets of the universal set (specific events) are represented as the interior of other geometric shapes – usually circles – inside the rectangle. Example on overhead…

Chapter 5 – Rules of Probability Complement of an Event – The denoted by A′ and consists of everything in the sample space S except event A. Since A and A′ together comprise the entire sample space, P(A) + P(A′ ) = 1. The probability of A′ is found by P(A′ ) = 1 – P(A). For example, The Wall Street Journal reports that about 33% of all new small businesses fail within the first 2 years. The probability that a new small business will survive is: P(survival) = 1 – P(failure) = 1 –.33 =.67 or 67%

Chapter 5 – Rules of Probability Odds of an Event: The odds in favor of event A occurring is Odds are used in sports and games of chance. If the odds against event A are quoted as b to a, then the implied probability of event A is: For example, if a race horse has a 4 to 1 odds against winning, then P(win) is P(A) = P(win) = or 20%

Clickers Suppose you apply for a job and are told that your odds against your being offered the job are 3 to 1. What is the probability that you will offered the job? A = 25% B = 33% C = 67% D = 75%

Chapter 5 – Rules of Probability Union of Two Events: The union of two events consists of all outcomes in the sample space S that are contained either in event A or in event B or both (denoted A  B or “A or B”).  may be read as “or” since one or the other or both events may occur. Intersection of Two Events: The intersection of two events A and B (denoted A  B or “A and B”) is the event consisting of all outcomes in the sample space S that are contained in both event A and event B.  may be read as “and” since both events occur. This is a joint probability.

Chapter 5 – Rules of Probability General Law of Addition: The general law of addition states that the probability of the union of two events A and B is: P(A  B) = P(A) + P(B) – P(A  B) Special Law of Addition: Events A and B are mutually exclusive (or disjoint) if their intersection is the null set (  ) that contains no elements. In the case of mutually exclusive events, the addition law reduces to: P(A  B) = P(A) + P(B)

Clickers Suppose we have two events, A and B, whose probabilities are P(A) = 0.7 and P(B) = 0.5. If P(A  B) = 0.3, find P(A  B). A = 0.3 B = 0.8 C = 0.9 D = 1.2

Chapter 5 – Rules of Probability Collectively Exhaustive Events: Events are collectively exhaustive if their union is the entire sample space S. Two mutually exclusive, collectively exhaustive events are dichotomous (or binary) events. For example, a car repair is either covered by the warranty (A) or not (B). Warranty No Warranty

Chapter 5 – Rules of Probability Collectively Exhaustive Events: More than two mutually exclusive, collectively exhaustive events are polytomous events. For example, a Wal-Mart customer can pay by credit card (A),debit card (B), cash (C) or check (D). Credit Card Debit Card Cash Check

Chapter 5 – Rules of Probability Forced Dichotomy: Polytomous events can be made dichotomous (binary) by defining the second category as everything not in the first category. For example… Polytomous Events Binary (Dichotomous) Variable Vehicle type (SUV, sedan, truck, motorcycle) X = 1 if SUV, 0 otherwise

Chapter 5 – Rules of Probability Conditional Probability: The probability of event A given that event B has occurred. Denoted P(A | B). The vertical line “ | ” is read as “given.” Consider the Venn Diagram… for P(B) > 0 and undefined otherwise

Chapter 5 – Rules of Probability Example – High School Dropouts: Of the population aged 16 – 21 and not in college: Unemployed13.5% High school dropouts29.05% Unemployed high school dropouts 5.32% What is the conditional probability that a member of this population is unemployed, given that the person is a high school dropout? First define: U = the event that the person is unemployed D = the event that the person is a high school dropout P(U) =.1350 P(D) =.2905 P(U  D) =.0532 or 18.31%

Clickers Suppose we have two events, A and B, whose probabilities are P(A) = 0.7 and P(B) = 0.5. If P(A  B) = 0.2, find P(A | B). A = 0.3 B = 0.4 C = 0.5 D = 0.7

Independent Events Independent Events: Event A is independent of event B if the conditional probability P(A | B) is the same as the marginal probability P(A). To check for independence, apply this test: If P(A | B) = P(A) then event A is independent of B. Another approach… Consider the definition of a conditional probability If A and B are independent, P(A|B) = P(A). So …

Independent Events Independent Events: Two events, A and B are independent if and only if P(A  B) = P(A)P(B). Generally, A collection of several events, E 1, E 2, E 3, …, E k, are mutually independent if and only if P(E 1  E 2  …  E k ) = P(E 1 )P(E 2 )…P(E k ).

Independent Events Dependent Events: When P(A) ≠ P(A | B), then events A and B are dependent. For dependent events, knowing that event B has occurred will affect the probability that event A will occur. Statistical dependence does not prove causality.

Clickers Suppose we have two independent events, A and B, whose probabilities are P(A) = 0.3 And P(B) = 0.4. Find P(A  B). A = 0.58 B = 0.64 C = 0.68 D = 0.70

Clickers Suppose we have two events, A and B, with probabilities are P(A) = 0.3, P(B) = 0.4, and P(A  B) =0.10. Are A and B independent? A = Yes B = No C = Not enough information is given.

Clickers True/False: If two events are mutually exclusive, then they are independent. A = True B = False

Independent Events The Five Nines Rule: How high must reliability be? Public carrier-class telecommunications data links are expected to be available % of the time. –The five nines rule implies only 5 minutes of downtime per year. If these links perform independently and have a probability of being available at a given time of 95%, how many redundant data links are necessary? System must be unavailable with probability < = For each independent link, p = P(unavailable) = = So, P(n redundant links unavailable) = p n.