© Buddy Freeman, 2015Probability

Segment 2 Outline  Basic Probability  Probability Distributions

© Buddy Freeman, 2015 Segment 2 Outline  Basic Probability  Probability Distributions Concepts of Probability Definitions Notation and Properties Complement Mutually Exclusive Collectively Exhaustive Addition Rules Conditional Probability Statistical Independence Multiplication Rules Discrete Probability Distributions General Concepts Specific Discrete Probability Distributions Hypergeometric Binomial Poisson Continuous Probability Distribution Normal

© Buddy Freeman, 2015 Concepts of probability 1) Classical - based on logic. 2) Relative Frequency - based on proportions. 3) Subjective - based on intuition and experience. 1) and 2) are quantitatively based and are emphasized here.

© Buddy Freeman, 2015 D E F I N I T I O N S Experiment: The experiment is the activity (process) that generates the possible outcomes (elements). It is what it is defined to be. Sample space: The set of all possible outcomes (elements) of the experiment. Event: An event is a subset of the sample space. An event contains zero or more of the outcomes of an experiment. Null event: The null event is the event with no outcomes. Certain (Sure) event: The event equal to the sample space.

© Buddy Freeman, 2015 Notation and Properties P(E): P(E) stands for the probability of event E. P(E) = (the number of outcomes in event E) (the number of outcomes in the sample space) P(E) = 0 iff E is the null event. P(E) = 1 iff E is equivalent to the sample space. For all events E, 0  P(E)  1.

© Buddy Freeman, 2015COMPLEMENT _ Complement: The complement of event E is denoted E. _ E is the set of all outcomes in the sample space that are NOT in event E. _ _ _ P(E) + P(E) = 1; P(E) = 1 - P(E); P(E) = 1 - P(E).

© Buddy Freeman, 2015 More Definitions Mutually exclusive: Two events, A and B, are mutually exclusive if they have no outcomes in common. (disjoint) Collectively exhaustive: A set of events is collectively exhaustive if it "fills up" ("covers") the sample space. Another way to define it is that the union of the events in the set contains the sample space. (A or B): The event that consists of all outcomes in event A, event B, or in both events A and B. (A and B): The event that consists only of the outcomes in event A that are also in event B. P(A and B) is sometimes called the joint probability of A and B.

© Buddy Freeman, 2015 Addition Rules General Addition Rule: Given any two events, A and B, P(A or B) = P(A) + P(B) - P(A and B). Special Addition Rule: Given two events, A and B, that are mutually exclusive P(A or B) = P(A) + P(B). (If A and B are mutually exclusive then the event (A and B) is null and P(null) = 0.) AB AB

© Buddy Freeman, 2015 Conditional Probability and Statistical Independence Conditional Probability: The probability of A given B is denoted P(A|B). If we assume that the event B has occurred, then what is the P(A) under this assumption? Formally, P(A|B) = P(A and B) where B is not null. P(B) Statistical Independence: Two events, A and B, are statistically independent iff P(A|B) = P(A). Equivalent definitions include: P(B|A) = P(B) or P(A and B) = P(A)  P(B).

© Buddy Freeman, 2015 Multiplication Rules General Multiplication Rule: Given two nonnull events, A and B, P(A and B) = P(A|B)  P(B) [= P(B and A) = P(B|A)  P(A)]. Special Multiplication Rule: Given two statistically independent events, A and B, P(A and B) = P(A)  P(B).

© Buddy Freeman, 2015 EXAMPLE PROBLEMS Experiment: Draw 1 card from a well-shuffled deck of 52 ordinary playing cards. (1) What is the probability that it is a Queen? (2) What is the probability that it is the Queen of Hearts (i.e. it is a Queen AND a Heart)? (3) What is the probability that it is a Queen OR a Heart? (4) If the selected card is a Heart, what is the probability that it is a Queen?

© Buddy Freeman, 2015 Problems continued... (5) Are the two events: drawing a Queen, and drawing a Heart mutually exclusive? (6) Are the two events: drawing a Queen, and drawing a Heart collectively exhaustive? (7) Are the two events: drawing a Queen, and drawing a Heart statistically independent?

© Buddy Freeman, 2015 More Example Problems Experiment: Draw 2 cards successively and without replacement from a well-shuffled deck of 52 ordinary playing cards. 1. What the the probability that both are Hearts? 2. If the first card is a Heart, what the the probability that the second card is a Queen? 3. What is the probability that the second card is a Queen? PROBLEM Prove or disprove: The two events: first card is a Heart and second card is a Queen are statistically independent.

© Buddy Freeman, 2015 Example Problem Handouts

© Buddy Freeman, 2015 Conditional Probability (e.g. Bayes Rule) Problems

© Buddy Freeman, 2015 Example 1 1. The BOZO steroid test was developed at DIZZY DEAN CHEMICALS, Inc.. The company claims that the BOZO steroid test is more than 90% effective both ways. If a subject has been using steroids, the test will indicate a positive result 92% of the time. If the subject does not use steroids, the test will indicate a negative result 95% of the time. It is estimated that the proportion of steroid users in the population is.002. If a subject is randomly selected and tests positive, what is the probability that this subject has been using steroids?

© Buddy Freeman, 2015 Example 2 2. A maker of personal computers buys memory chips through three different suppliers which shall be referred to as A, B, and C. Twenty percent of their supply comes from A, fifty percent from B, and the remainder from C. It is known that the proportion of defective chips produced by the three suppliers are.0010,.0003, and.0007, respectively. (a) If one of the memory chips is defective, which supplier is most likely to have supplied it? (b) What is the probability of a defective chip?

© Buddy Freeman, 2015Factorials Factorials: 0! = 1; 1! = 1; 2! = 2  1; 3! = 3  2  1; in general n! = n(n-1)(n-2)...  3  2  1

© Buddy Freeman, 2015 Counting Rules 1) When order matters, the following are useful. (a) If it can be done n 1 ways at the first trial, n 2 ways at the second trial,..., and n m ways at the m th trial the total number of ways is n 1  n 2 ...  n m. (b) When sampling without replacement from n distinct objects we can select m (where m  n) objects n!/(n-m)! ways. These are commonly referred to as permutations of size m taken from a set of size n (denoted n P m ). 2) When order does not matter, you divide by the number of ways that you can order the m objects that you are selecting. Thus, when sampling without replacement from n distinct objects we can select m (where m  n) objects n!/[(n-m)!(m)!] ways. These are commonly referred to as combinations of size m taken from a set of size n.