Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
The probability model The probability model Math 425 Introduction to Probability Lecture 5 + The basis of probability is the experiment: An experiment is a repeatable procedure that has a a measurable outcome which cannot be predicted ahead of time. + The probability model consists of three ingredients Kenneth Harris [email protected] 1 A sample space S of possible outcomes of an experiment, 2 A collection E of events, which are subsets of S, and are said to occur as the outcome of an experiment. 3 A probability measure P which assigns a nonnegative real number to events (the “probability of the event occurring as the outcome of an experiment"). Department of Mathematics University of Michigan January 19, 2009 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 1/1 Kenneth Harris (Math 425) The probability model Math 425 Introduction to Probability Lecture 5 January 19, 2009 3/1 The Axioms of Probability Events Enter Tyche + Which subsets of the sample space S are events? + A probability measure on a sample space S is a function P defined on the events E of the space S. We write Discrete sample space. In a discrete sample space, we can safely take any subset of the sample space S to be an event, which has some probability of occurring. P (E) Nondiscrete sample space. Not all subsets of a nondiscrete sample space can be events (which have some probability of occurring). This issue will never be a problem for us in this class. It is actually hard to construct such events in the continuous sample spaces in which we will be most interested. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 4/1 to mean the probability of the event E occuring Sample space. Whenever we talk about a probability measure P , we are assuming it is a measure with respect to some given sample space S. + A probability measure must also satisfy the three axioms which follow. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 6/1 The Axioms of Probability The Axioms of Probability Axiom 1 Axiom 2 Axiom (1 – Nonnegative) Axiom (2 – Unit measure) For any event E, P (E) is a real number and P (S) = 1. 0 ≤ P (E). + This axiom simply states that P (E) is a nonnegative real number for any event E. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 7/1 + The certain event S must always occur, and is defined to have probability 1. Kenneth Harris (Math 425) The Axioms of Probability Math 425 Introduction to Probability Lecture 5 Extended Addition Law Axiom (3 – Addition rule (weak form)) There is nothing special about two events. For any mutually exclusive events E and F (so, E ∩ F = ∅), Proposition (Extended Addition Law) If events E1 , E2 , . . . , En are mutually exclusive (so, Ei ∩ Ej = ∅ whenever i 6= j), then P (E ∪ F ) = P (E) + P (F ). P (E1 ∪ E2 ∪ . . . ∪ En ) = P (E1 ) + P (E2 ) + . . . + P (En ). + This version of Axiom 3 is most relevant when the sample space S is finite. Compare this axiom to the Sum Rule for counting: In shorthand, P If two events E and F are mutually exclusive and E has n outcomes and F has m outcomes, then E ∪ F has n + m outcomes. Math 425 Introduction to Probability Lecture 5 8/1 The Axioms of Probability Axiom 3 – Weak form Kenneth Harris (Math 425) January 19, 2009 January 19, 2009 9/1 Kenneth Harris (Math 425) n [ n X Ei = P (Ei ). i=1 i=1 Math 425 Introduction to Probability Lecture 5 January 19, 2009 10 / 1 The Axioms of Probability The Axioms of Probability Proof Axiom 3 – Strong form I’ll prove the case of n = 3. Let E, F and G be mutually exclusive events. Then E ∪ F and G are also mutually exclusive. (Why?) Axiom (3 – Addition rule (strong form)) For any sequence of mutually exclusive events E1 , E2 , . . . (so, Ei ∩ Ej = ∅ whenever i 6= j), + Apply the Addition Rule (Axiom 3): P E ∪ F ∪ G = P (E ∪ F ) ∪ G P( = P (E ∪ F ) + P (G) ∞ [ Ek ) = k =1 ∞ X P (Ek ). k =1 = P (E) + P (F ) + P (G) + This rule is only relevant when the sample space S is infinite. The Strong form of the Addition Rule really is stronger than its Weak form. (See Ross, p. 30). + The n-event case is proved by induction, in the same way. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 11 / 1 Kenneth Harris (Math 425) Consequences of the Axioms Math 425 Introduction to Probability Lecture 5 January 19, 2009 12 / 1 January 19, 2009 15 / 1 Consequences of the Axioms Probability of the Impossible Event Proposition 4.1 Recall. ∅ is the impossible event. Proposition (Ross 4.1) For any event E, Proposition P (E c ) = 1 − P (E). P (∅) = 0. Proof. Both S = E ∪ E c and E ∩ E c = ∅. So, Proof. Both S = S ∪ ∅ and S ∩ ∅ = ∅. Apply the Addition Rule (Axiom 2): 1 = P (S) = P (E ∪ E ) = P (E) + P (E c ) P (S) = P (S ∪ ∅) = P (S) + P (∅). Axiom 3 So, P (E c ) = 1 − P (E). So, P (∅) = 0. Kenneth Harris (Math 425) Axiom 2 c Math 425 Introduction to Probability Lecture 5 January 19, 2009 14 / 1 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 Consequences of the Axioms Consequences of the Axioms Proposition 4.2 Proof of Proposition 4.2 Proof. Since E ⊆ F , we can express F as (see next slide) Proposition (Ross, 4.2) F = E ∪ (E c ∩ F ) Let E and F be any events. If E ⊂ F then P (E) ≤ P (F ). where E and E c ∩ F are mutually exclusive. By the Addition Rule As a consequence, 0 ≤ P (E) ≤ 1 for any event E. P (F ) = P (E) + P (E c ∩ F ) ≥ P (E), where the last is because P (E c ∩ F ) ≥ 0 by Axiom 1. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 16 / 1 Kenneth Harris (Math 425) Consequences of the Axioms Math 425 Introduction to Probability Lecture 5 January 19, 2009 17 / 1 Consequences of the Axioms Partition for Proposition 4.2 Proposition 4.3 Principle. Let E and F be any events in the same sample space. Then E = E ∩ F and E c ∩ F are mutually exclusive, and F = (E ∩ F ) ∪ (E c ∩ F ) = E ∪ (E c ∩ F ). + The following is the simplest version of the Inclusion-Exclusion Identity. (See Ross, Proposition 4.4. This is for a later lecture.) S F Proposition (Ross 4.3) E Let E and F be any events. Then EÝF P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ) Ec Ý F Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 18 / 1 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 19 / 1 Consequences of the Axioms Consequences of the Axioms Proof of Proposition 4.3 Partition for Proposition 4.3 + By the Addition Rule (see next slide) Principle. Let E and F be any events in the same sample space. Then each of E ∩ F , E c ∩ F and E ∩ F c are mutually exclusive, and P (E ∪ F ) = P (E ∪ F c ) + P (E c ∪ F ) + P (E ∩ F ). E ∪ F = (E ∩ F ) ∪ (E c ∩ F ) ∪ (E ∩ F c ). S + On the other hand, the Addition Rule gives (see next slide) P (E) = P (E ∪ F c ) + P (E ∩ F ) P (F ) = P (E c ∪ F ) + P (E ∩ F ) E Ý Fc EÝF Ec Ý F So, E P (E) + P (F ) − P (E ∩ F ) = P (E ∪ F c ) + P (E c ∪ F ) + 2 · P (E ∩ F ) = P (E ∪ F c ) + P (E c ∪ F ) + P (E ∩ F ) = P (E ∪ F ) F −P (E ∩ F ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 20 / 1 Kenneth Harris (Math 425) Consequences of the Axioms Math 425 Introduction to Probability Lecture 5 January 19, 2009 21 / 1 Consequences of the Axioms Example 1 Example 1 – solution Solution. Consider the events: A: customers who carry AMEX, V : customers who carry VISA. Example A retail store accepts VISA and AMEX. It has found that 24 percent of its customers carry AMEX, 61 percent carry VISA and 11 percent carry both. What percentage of its customers carry VISA or AMEX? Then, from the data P (A) = 0.24 P (V ) = 0.61 P (A ∩ V ) = 0.11 Apply Proposition 4.4, P (A ∪ V ) = P (A) + P (V ) − P (A ∩ V ) = 0.24 + 0.61 − 0.11 = 0.74. + So, 74 percent of the customers carry a credit card. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 22 / 1 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 23 / 1 Consequences of the Axioms Consequences of the Axioms Example 2 Example 2 – solution Solution. Consider the events: G: students who are Greek I: students involved in intramural sports. Example Then, from the data (since Gc ∩ I c = (G ∪ I)c ) Sixty percent of students at a certain school are neither Greek nor participate in an intramural sports. Twenty percent are Greek and 30 percent participate in intramural sports. P (G ∪ I)c = 0.6 P (G) = 0.2 P (I) = 0.3 (a). By Proposition 1.1 (a) What is the percentage of students who are Greek or participate in intramural sports? (b) What is the percentage of student who do both? P G ∪ I = 1 − P (G ∪ I)c = 1 − 0.6 = 0.4 So, (a) 40 percent of student are either Greek or participate in intramural sports. (b). Apply Proposition 4.3, P (G ∩ I) = P (G) + P (I) − P (G ∪ I) = 0.2 + 0.3 − 0.4 = 0.1 So, (b) 10 percent of student participate in both activities. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 24 / 1 Finite sample space with equally likely outcomes Kenneth Harris (Math 425) Example 1 + Let S is a finite sample space, and suppose it has N outcomes, listed as S = {a1 , a2 , . . . , aN } Example + Then it follows from the Axioms that: Solution. There are 36 possible outcomes on a pair of dice (any number 1 to 6 could appear on each die). The throws with the same face are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). For every i ≤ N, 1 P {ai } = , N For any event E, + The desired probability is |E| , N where |E| is the number of outcomes in E. P (E) = Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 25 / 1 Assume all outcomes of a throw of a pair of dice are equiprobable. What is the probability of rolling the same value on both die? + Suppose every outcome is equiprobable: P {a1 } = P {a2 } = . . . = P {aN } 2 January 19, 2009 Finite sample space with equally likely outcomes Finite sample spaces, equiprobable outcomes 1 Math 425 Introduction to Probability Lecture 5 January 19, 2009 27 / 1 Kenneth Harris (Math 425) 6 36 = 16 . Math 425 Introduction to Probability Lecture 5 January 19, 2009 28 / 1 Finite sample space with equally likely outcomes Finite sample space with equally likely outcomes Example 2 Example 2 – solution Solution. There are 25 possible outcomes. + Let Ei be the event that the ith toss is heads. Then |Ei | = 24 (fix ith toss at H). So, 24 1 P (Ei ) = 5 = . 2 2 Example A coin is tossed five times in succession. Assume each sequence of heads/tails is equiprobable. What is the probability of getting heads on the first or third toss? + For i 6= j, |Ei ∩ Ej | = 23 (fix ith and j toss at H). So, P (Ei ∩ Ej ) = + The sample space S for this problem are sequences (a1 , a2 , a3 , a4 , a5 ) 1 23 = . 5 2 4 + Apply Proposition 4.4: where each ai = H or T P (E1 ∪ E3 ) = P (E1 ) + P (E3 ) − P (E1 ∩ E3 ) = For example: (H, T , H, T , H). 1 1 1 3 + − = . 2 2 4 4 + The desired probability is 34 . Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 29 / 1 Finite sample space with equally likely outcomes Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 30 / 1 Finite sample space with equally likely outcomes Example 3 Example 3 – solution Example Solution. The sample space of the Pick-Six Lottery are all subsets of {1, 2, 3, . . . , 49} with six elements: 49 possible outcomes . 6 Some states have the Pick-Six Lottery: A person purchases a ticket, and can choose 6 distinct numbers in the set {1, 2, 3, . . . , 49}. Later a Lottery Machine picks 6 distinct numbers at random in the set {1, 2, 3, . . . , 49}. A winning ticket is one which matches the six numbers chosen by the Machine (in any order of selection). + By assumption, all possible selections of numbers are equiprobable, so the likelihood that a ticket is a winner is 1 = 49 6 + What are the odds of winning with one ticket? 1 . 13, 983, 816 + You have better odds getting 23 consecutive tosses of heads on a fair coin (where heads/tails are equally likely on each toss). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 31 / 1 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 5 January 19, 2009 32 / 1