1 Probability Ch2

2   The Universal Set, denoted S, is the set of all objects under consideration at a given moment. Also known as the outcome space or the population or the sample space.   An Element of a set is any member of the set.   A Subset is any part of a set; A B means A is a subset of B.   An Event is a collection of possible outcomes and must be contained in S.   The Empty (Null) Set is a notational idea representing the set (or subset) with no members; we use Ø to denote it.

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4   The basic set operations: Union, Intersection, Complement, Set Difference.   A  B is the union of A and B; A  B is the intersection of A and B;   A’ is the complement of A; A − B is the difference of A and B.   Two sets are Mutually Exclusive or disjoint if they have no elements in common (their intersection is disjoint).   A 1,A 2,...,A k are mutually exclusive events means that A i  A j = Ø  , i  j, that is, A 1,A 2,...,A k are disjoint sets.   A 1,A 2,...,A k are exhaustive events means A 1  A 2 ...  A k = S.   Venn diagrams are useful is graphically representing sets. Laws on Set operations.   A Set Function is a function whose domain is a collection of sets.

5   The events and outcome space are sets.   Fundamental set operations can be applied in the probability computation.   They are ∩, ∪, and the like.   For event A in an experiment of n trials,   The occurrences of A are denoted as N(A), A's frequency.   N(A)/n is the relative frequency of A.A's probability P(A), as n↑.   Ex2.1-1: rolling a dice 6 times–success: there is a match if dice=k on Roll k. Let A = {success}. P(A)=1-(1-1/6) 6 = N(A)/n can be verified by simulation, and will approach.665 as n↑.   Ex2.1-2: throwing a disk with diameter 2 inches on a tiled floor, where each tile is a square with sides 4 inches in length. Let C = {the disk is placed entirely on one tile.}. P(C) = 4/16 = 0.25.

6 Definition and Theorem   Def.1.2-1: Probability is a real-valued set function P that assigns to each event A.the sample space S a number P(A), called the probability of event A.   P(A) ≥0, P(S)=1   If A 1, A 2, …, A k are disjoint events, then P(A 1 ∪ A 2 ∪ … ∪ A k ) = P(A 1 ) + P(A 2 ) + …+ P(A k ) for any integer k.   Thm1.2-1: For each event A, P(A) = 1 –P(A’).   Thm1.2-2: P(  )=0.   Thm1.2-3: If event A  event B, then P(A) ≤P(B).   Thm1.2-4: For each event A, P(A) ≤1.

7   Thm1.2-5: For any two events A & B, P(A ∪ B) = P(A) + P(B) –P(A ∩B).   Thm1.2-6: For any three events A, B & C, P(A ∪ B ∪ C) = P(A) + P(B) + P(C) –P(A ∩B) –P(B ∩C) –P(C ∩A) + P(A ∩B ∩C)   Ex1.2-6: Draw a card at random from a deck of 52 cards. Let A denote the outcome is a king. Then, P(A)=4/52=1/13.

8 Enumeration Methods   Multiplication Principle:   An experiment E 1 has n 1 outcomes, and for each of them, an experiment E 2 has n 2 outcomes.   The composite experiment E 1 E 2 for doing first E 1 and then E 2 has n 1 n 2 outcomes.   Ex2.2-1: E 1 denotes the rat selection from a cage –{F, M}, and E 1 denotes applying drug to the selected rate –{A, B, P}. The set of all possible composite outcomes is {(F, A), (F, B), (F, P), (M, A), (M, B), (M, P)}, 6 in total.   Tree diagram:   The composite experiment E 1 E 2 …E m has n 1 n 2 …n m outcomes (different combinations).

9 Definition   Def.1.3-1: Each of the n! arrangements (in a row) of n different objects is called a permutation of the n objects.   There are n positions to be filled with n different objects  n!   Ex2.2-3: The number of permutations of the 4 letters a, b, c, and d clearly 4!=24. Particularly, the number of possible 4-letter code words using these 4 letters is 4 4 =256 if letters can be repeated (reused).   Def.1.3-2: Each of the n P r arrangements is called a permutation of n objects taken r at a time.   n P r = n(n-1)(n-2)…(n-r+1)=n!/(n-r)!   Ex1.3-5: The number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club with 10 persons is 10P4=5040.   These 4 persons chosen are distinguishable.   Def.1.3-3: If r objects chosen from a set of n objects are distinguishable or the selection order is noted, the selected set of r objects is called an ordered sample of size r.

10 Definition (cont'd)   Def.1.3-4: Sampling with replacement occurs when an object is selected and then replaced before the next object is selected.   The no. of possible ordered samples of size r taken from n objects is n r.   Ex1.3-6: Rolling a dice 5 times. The number of possible ordered samples is 6 5 =    Rolling a dice is equivalent to sampling with replacement.   Def.1.3-5: Sampling without replacement occurs when an object is not replaced after it has been selected.   There are exactly n P r possible ordered samples.   If the selection order of r objects is ignored, the number of (unordered) subsets of size r is n C r = n P r /r! = n!/[r!(n-r)!].

11   Def.1.3-6: Each of the n C r unordered subsets is called a combination of n objects taken r at a time.

12 Example   Ex1.3-11: Drawing a five-card hand from a deck of 52 cards.   Event A: the five cards in a hand are all spades.   Event B: there are exactly three kings and two queens.   Event C: there are exactly two kings, two queens, and one jack.

13 Distinguishable Permutations   Def.1.3-7: Each of the n C r permutations of n objects, r of one type and n-r of another type, is called a distinguishable permutation.   Ex1.3-12: Flipping a coin 10 times to yield a head/tail sequence. The number of possible outcomes having 4 heads and 6 tails is   Ex1.3-13: Arranging 7 colored flags on a vertical flagpole. The number of different signals by 4 orange and 3 blue flags is   Extension: For a set of n objects, n 1 are similar, n 2 are similar, …, n s are similar, where n 1 + n 2 + …+ n s = n. The number of distinguishable permutations of the n objects is  multinomial coefficients

14 Selecting r objects out of n objects   For ordered samples,   there are n r possible outcomes when sampling with replacement.   Each of r different objects is arbitrarily mapped to one of n objects.   there are n P r outcomes when sampling without replacement.   For unordered samples (the order is irrelevant),   there are n C r outcomes when sampling without replacement.   there are n-1+r C r outcomes when sampling with replacement.   This can be computed as the number of ways to insert (n-1) “|”into a series of r “0”.   For instance, 00||000|0|000|0 is an example for n=6 and r=10.   Each of r similar objects is arbitrarily mapped to one of n objects.

15   Ex1.3-15: Rolling a pair of six-sided dices. The number of distinguishable outcomes is   Note: each outcome occurs unequally likely!

16 More Examples   Ex: Giving a, a, a, a, b, c, d, e arbitrarily to 3 persons.   The number of distinguishable outcomes is H 3 4  3 4   Ex: Giving a, b, c, d, e, f items arbitrarily to   4 persons, each with exactly one item.  The number is P 6 4   10 persons, each with at most one item.  The number is   4 persons.  The number is   4 persons, but the first person gets at least 2 items.  The number is   Ex: Giving a, a, a, a, a, a items arbitrarily to   4 persons, each with exactly one item.  The number is 1.   10 persons, each with at most one item.  The number is C 10 6   4 persons.  The number is H 4 6   4 persons, but the first person gets at least 3 items.  The number is H 4 3   (As giving a, a, a items arbitrarily to 4 persons.)

17 Conditional Probability   Def.1.4-1: P(A|B) is a conditional probability, defined as the probability of A among the occurrences of B.   Mathematically, the relationship is   Note: P(B)>0.   Ex.1.4-2: Suppose P(A)=0.4, P(B)=0.5, and P(A∩B)=0.3.  P(A|B)=0.3/0.5=0.6; P(B|A)=0.3/0.4=0.75

18   Ex.1.4-4: Rolling a pair of four-sided dice and determine the sum. Let A be the event that a sum of 3 is rolled, and B be the event that a sum of 3 or 5 is rolled. Compute P(A|B).   P(A|B): a sum of 3 occurs before a sum of 5.

19 Conditional Probability  Derivation   With P(B)>0, Def can be specified for conditional probability.   P(A|B) ≥0, P(B|B)=1.   If A 1, A 2, …, A k are disjoint (mutually exclusive) events, then P(A 1 ∪ A 2 ∪ … ∪ A k |B) = P(A 1 |B) + P(A 2 |B) + …+ P(A k |B) for any integer k.   Ex1.4-5: There are 25 balloons: 10 yellow, 8 red, and 7 green. Given the balloon sold is yellow, what is the probability that the next balloon selected at random is also yellow?  This conditional probability is 9/24.   A: the 1 st balloon sold is yellow; B: the 2 nd balloon selected is yellow. P(B|A)=9/24=P(A∩B)/P(A).P(A∩B)= P(B|A)P(A)=(9/24)(10/25).

20   Def.1.4-2: The probability that two events, A and B, both occur is given by the multiplicative rule P(A∩B)= P(B|A)P(A)=P(A|B)P(B).   Individual probabilities, P(A), P(B), P(B|A), P(A|B), may be easier to get. [an factorization idea: divide/conquer]

21 Examples   Ex.1.4-6: Drawing 2 out of 10 chips: 7 blue and 3 red. What is the probability that (A) the first drawn is red, and (B) the second drawn is blue?   It is easier to get P(A)=3/10 and P(B|A)=7/9. Thus P(A∩B)=7/30.   Ex.1.4-7: Drawing cards at random w/o replacement. What is the probability the third spade appears on the sixth draw?   A: two replacement spades in the first five draws.   B: a spade is drawn on the sixth draw.   Different factorizations are possible, if reasoning is consistent with assumptions.

22   Ex.1.4-8: Rolling a pair of 4-sided dice. What is the probability for rolling a sum of 3 on the first roll and then, rolling a sum of 3 before rolling a sum of 5? Ex.1.4-4) (2/16) (2/6)=1/24

23 Chained Conditional Probability   P(A  B  C)=P(A  B)P(C|A  B) =P(A)P(B|A)P(C|A  B)   Ex1.4-9: Drawing 4 cards from an ordinary deck. The probability of receiving in order a spade, a heart, a diamond, and a club is (13/52)(13/51)(13/50)(13/49)   Ex1.4-10: The left pocket has 5 blue and 4 white marbles and the right pocket has 4 blue and 5 white. If one marble is transferred from left to right, then the probability of subsequently drawing a blue marble from the right is   BL: drawing a blue from the left.   BR: right.   WL: white left. P(BR)=P(BL  BR)+P(WL  BR) =P(BL)P(BR|BL)+P(WL)P(BR|WL) =(5/9)(5/10)+(4/9)(4/10)=41/90

24 Independent Events   If the occurrence of one event does not change the probability of the other's happening, both are said to be independent events.   Def.1.5-1: Events A & B are independent iff P(A∩B)=P(A)P(B). Otherwise, they are called dependent events.   Ex1.5-1: Flip a coin twice; the sample space of the sequence is S={HH, HT, TH, TT}, each with probability of 1/4.   A={heads on the first flip}={HH, HT}   B={tails on the second flip}={HT, TT}   C={tails on both flips}={TT}   P(B)=1/2; P(B|C)=1 since C  B.  C and B are dependent.   P(B|A)=1/2=P(B).  A and B are independent.  P(A|B)=P(A).   P(A∩B)=P(A)P(B).   Ex1.5-2: Roll a red die and a white die. A={4 on red}; B={sum is odd}   P(A)P(B)=(6/36)(18/36)=3/36=P(A  B) Independent   Ex1.5-3: C={5 on red}; D={sum is 11}  P(C)P(D)=(6/36)(2/36)=1/108  1/36=P(C  D) Dependent

25 Derived Independences   Thm1.5-1: If A & B are independent, the following are too:   A & B'; A' & B; A' & B'.   Ex1.5-4: Draw a ball from an urn with 4 balls: 1, 2, 3, 4.   A={1, 2}, B={1, 3}, C={1, 4}.  P(A)=P(B)=P(C)=1/2.   A∩B={1}, B∩C={1}, A∩C={1}.  P(A∩B)=P(B∩C)=P(C∩A)=1/4. ∵ P(A∩B)= P(A)P(B), … ∴ A, B, C are independent in pairs. (pairwise independence)   However, P(A∩B∩C)=P({1})=1/4 ≠1/8=P(A)P(B)P(C).  There is no complete independence of A, B, and C. P(A  B’)=P(A)P(B’|A) =P(A)[1-P(B|A)] =P(A)[1-P(B)]=P(A)P[B’]

26   Def.1.5-2: A, B & C are mutually independent iff   They are pairwise independent, and   P(A∩B∩C)=P(A)P(B)P(C).   Generally, each pair, triple, quartet, etc. must be all independent.    A', B' & C' are mutually independent;    A and (B∩C); A and (B ∪ C); A' and (B∩C'); …are independent.

27 Representation of Sample Space   Ex1.5-6: Rolling a 6-sided dice 6 times.   A i : Side i is observed on Roll i –a match.  P(A i )=1/6 & P(A i ')=5/6.   B: at least one match occurs; B': no matches occur.  P(B)=1-P(B')=1-P(A1'∩…∩A6')=1-(5/6) 6.   Roll a 6-sided die n times.   S = {(O 1, …, O n ): O i =1,2,…,6, for i = 1,2,…,n}.   Toss a coin n times.   S = {(O 1, …, O n ): O i =H or T, for i = 1,2,…,n}.   For simplicity, (O 1, …, O n ) = O 1 …O n ; (H, T) = HT.   Ex1.5-9: On 5 consecutive days, a lottery with the winning probability of 1/5 is purchased on each day.   Assume independence, P(WWLLL)=(1/5) 2 (4/5) 3 =P(LWLWL).   The probability of purchasing 2 winning tickets and 3 losing tickets is

28 Bayes' Theorem   Posterior Probability of B k by some A! Note: B 1, …, B m are mutually exclusive and exhaustive.   The prior probability of event Bi: P(B i )>0, for i=1..m.

29   Ex.1.6-2: Machines A, B, C are all producing springs of the same length.   Of the total productions, they produce 35, 25, and 40%, with the defective rates of 2, 1, and 3%, respectively.   The probability of selecting a spring that is defective is P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) = 35% 2% + 25% 1% + 40% 3% = 215/   The conditional probability that the defective spring is produced from C is