Chapter Probability part 1 Experiment - you do something or measure something and note the outcome. Random experiment - the outcome isn't always the same. Basic Concepts Sample space, of an random experiment - the set of all possible outcomes, denoted by S. Null set empty set, ={ } Event a subset of the sample space: E S Each outcome is an event is an event, impossible event S is an event, certain event Examples: Continuous: Recycle time of a flash S={x x>0} Discrete: Survey question: S={yes, now}; Grouped age: S={young, middle, old} Tree diagram Message delays Message 1: on time or late Message : on time or late Message 3: on time or late S={ooo, ool, olo, oll, loo, lol, llo,lll} Example -5 (p0): Automobile colors Exterior color: red, white, blue, brown Interior color: red(), w(4),b(3),b(1) Composite Events The union of two events E 1 E is another event that contains all the sample points that are in either one of the other two. "or The intersection of two events E 1 E is another event that contains all the sample points that are both of the other two. "and The complement, E, of an event is another event that contains all the other outcomes in the sample space that are not in the original event. "not" Mutually Exclusive Partition 1
Counting Techniques Multiplication Principle: Sequential operations of k steps. 1st step has n1 ways, nd step has n ways,, kth step has nk ways, then the total number of ways of completing the operation is n1*n* *nk. Example: Design of a casing for a gear housing: If there are 4 types of fasteners, 3 bolt lengths, and 3 bolt location, then there are 4*3*3=36 different possible designs. Permutations Permutation: The number of ordered sequences The number of permutation of n different elements =n*(n-1)*(n-)* **1 The number of permutation of r elements from n objects: P n r n*( n 1)*( n )*...( n r 1) ( n r)! The number of permutations of n=n1+n+ +nr objects: n! n!... n! 1 r Examples Example -11: A hospital needs to schedule 3 knee surgeries and hip surgeries in a day. The number of possible sequences is 5! 10 10 3!! 1 Combinations Combination: The number of subsets of r elements that can be selected from a set of n elements. The number of combinations of r elements from n elements is n C n r r r!( n r)! Examples Sampling without replacement: A bin of n=50 manufacture parts contains 3 defective parts and 47 non-defective parts. A sample of 6 parts is randomly selected without replacement. How many different samples of size 6 that contain exactly defective parts? Solution: It contains steps. Step 1: The number of ways choosing defective from 3 defectives: 3 3! 1!! 6 3
Solution Solution: It contains steps. Step 1: The number of ways choosing defective from 3 defectives: 3 3! 1!! 6 3 Step : The number of ways choosing the remaining 4 from 47 acceptable parts: 47 4 47! 178,365 4!43! By the multiplication principle, the number of subsets of size 6 that contains exactly defective parts is 3*178,365=535,095. How many different subsets of size 6 can be found? 50 15,890,700 6 What is the probability of getting a subset of size 6 that contains D only?. Probability A probability function, P, is a real function valued between zero and one. The integral over the entire domain (S) of a probability function = 1. Discrete probability function- the sample space, S, the domain of the function, is finite or countable infinite. Continuous probability function - the sample space contains an interval of the real numbers Axioms of probability S)=1 0E)1 E 1 E = E 1 E )=E 1 )+ E ) (*) (1) E )=1-E) () If E 1 E, then E 1 ) E ) Equally Likely Probability Model: (S,P) Each of the basic outcomes has the equal chance to be selected. Let S={e1,, en}. ei)=1/n, i=1,,n. Let A be an event containing #( basic outcomes. Then by (*), #( #( S) Example: Assume that 30% of the laser diodes in a batch of 100 meet the minimum power requirements of a specific customer. If a laser diode is selected randomly, that is, each laser diode is equally likely to be selected, then the probability of meeting the customer s requirements is 0.30. Example A random experiment can result in one of the outcomes {a,b,c,d,e} with probabilities 0.1,0., 0.,0.4, 0.1. The sample space is {a,b,c,d,e}. Probability Distribution Let A ={a,b} and B={c,d} Then = B)= A )= AB)= AB)= a b c d e 0.1 0. 0. 0.4 0.1 3
Example: Wafer Contamination Example: Manufacturing inspection # of contamination particles 0 0.40 1 0.0 0.15 3 0.10 4 0.05 5 or more 0.10 Proportion of Wafers Let E={0}. E)=0.40. at most 3 contaminated particles in the inspected location) =0)+1)+)+3) =0.40+0.0+0.15+0.10=0.85 at least 3 contaminated particles in the inspected location) =3)+4)+5 or more) =0.10+0.05+0.10=0.5 a sample of size 6 contains exactly defective) =535,095/15,890,700=0.034 a sample of size 6 contains no defective) =C(47,6)/C(50,6)=10,737,573/15,890,700=0.676. Two-way contingency table Example: Wafers in semiconductor manufacturing are classified by contamination and location. Relative Frequency Table Convert to Joint Probability table by dividing by the total: Location Contamination Center Edge Total Low 514 68 58 High 11 46 358 Total 66 314 940 514/940=.547, joint probability 3-D Graph of joint probability function Additive Rule For given two events, A and B, AB)=+B)-AB) 4
.4 Conditional Probability A B) B, 0. B given A Tree diagram and Conditional Probability Example (Surface flaws): 400 parts are classified by surface flaws (F) and as (functionally) defective (D). Conditional Probability has all properties of Probability, except the sample space is A. High Center)=High and Center)/Center) D F)=10/40 =(11/940)/(66/940)=11/66=0.179 Sampling without replacement -6 More Examples Use tree diagram: (a) 4/499 (b) (5/500)(4/499) by multiplication rule (c) (495/500)(494/499) 5