STAT 430/510 Probability Hui Nie Lecture 3 May 28th, 2009
Review We have discussed counting techniques in Chapter 1. Introduce the concept of the probability of an event. Compute probabilities in certain situations.
Experiment A random experiment is a process whose outcome is uncertain. Example: Tossing a coin once or several times; Picking a card or cards from a deck; Measuring temperature of patients;
Events and Sample Spaces A sample spaces S of a random experiment is the set of all possible outcomes. An event E is any subset of the sample space S. Our objective is to determine P(E), the probability that event E will occur.
Example The experiment: Toss a coin 3 times. Sample space S={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Examples of event include A={HHH,HHT,HTH,THH}={at least two heads} B={HTT,THT,TTH}={exactly two tails}
Example An experiment in a hospital consists of recording the sex of each newborn infant until the birth of a male is observed. The sample space of this experiment is S = {M, FM, FFM, FFFM, }. The sample spaces contains an infinite number of outcomes.
Example An executioner offers a death-row prisoner a final chance to gain his release. 20 chips, 10 white and 10 blue. All 20 are to be put into two urns by the prisoner with each contains at least one chip. The executioner will pick one urn randomly and from that urn, one chip at random. If the chip is white, the prisoner will be set free; if it is blue, he will be executed.
Example What is the sample space describing the prisoner s possible allocation options? S = {[(1, 0), (9, 10)], [(1, 1), (9, 9)],, [(0, 1), (10, 9)]} = {[(w 1, b 1 ), (w 2, b 2 )] : w 1 + b 1 > 0, w 2 + b 2 > 0, w 1 + w 2 = 10, b 1 + b 2 = 10, w 1, w 2, b 1, b 2 0} (Intuitively, what s the best allocation for the prisoner?)
Basic Concepts The union of two events A and B, A B, is the event consisting of all outcomes that are either in A or in B or in both events. The complement of an event A, A c, is the set of all outcomes in S that are not in A. The intersection of two events A and B, A B or AB, is the event consisting of all outcomes that are in both events. When two events A and B have no outcomes in common, AB =, they are said to be mutually exclusive, or disjoint, events.
Example The experiment: toss a coin 10 times and the number of heads is observed. Let A = {0, 2, 4, 6, 8, 10}, B = {1, 3, 5, 7, 9}, C = {6, 7, 8, 9, 10}. A B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = S AB =. So A and B are mutually exclusive. C c = {0, 1, 2, 3, 4, 5}, AC c = {0, 2, 4}
Rules Commutative Laws: A B = B A, AB = BA Associative Laws: (A B) C = A (B C), (AB)C = A(BC) Distributive Laws: (A B)C = AC BC, (AB) C = (A C)(B C) DeMorgan s Laws: ( n i=1 A i) c = n i=1 Ac i, ( n i=1 A i) c = n i=1 Ac i These laws can be shown by Venn diagram.
Probability Distribution Probabilities are values of a set function, also called a probability distribution. This function assigns real numbers to the various subsets (events) of a sample space S. Probability of an event can be interpreted as the limiting relative frequency of the event. Probability distributions satisfy the following axioms.
Axioms of Probability Axiom 1: 0 P(E) 1 Axiom 2: P(S)=1 Axiom 3: For any sequence of mutually exclusive events E 1, E 2,, (that is, events for which E i E j = when i j), P( E i ) = i=1 P(E i ) i=1
Interpretation These axioms of probability require no proof. The axioms only serve to exclude assignments inconsistent with our intuitive notion of probability.
Example A fair die is rolled. P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1 6 P({2, 4, 6}) = 1 2
Properties of Probability P(E c ) = 1 P(E) If E F, then P(E) P(F) P(E F) = P(E) + P(F) P(EF)
Example J is taking two books along on her holiday vacation. With probability 0.5, she will like the first book; with probability 0.4, she will like the second book; and with probability 0.3, she will like both books. What is the probability that she likes neither book? Let B i denote the event that J likes book i, i = 1, 2. P(B c 1 Bc 2 ) = P((B 1 B2 ) c ) = 1 P(B 1 B2 ) = 1 (P(B 1 ) + P(B 2 ) P(B 1 B 2 )) = 0.4