[확률과 통계] CH01 - Probability and Statistics

• Sample Space : all of the possible experimental outcomes
  • ex) chips in boxes of 500, each chips can be classified as either satisfactory or defective. 
  • the number of defective chips in bos is uncertain
    • S = {0 defectives, 1 defectives, ... , 500 defectives }
  • ex2) separate errors in a particular piece of software
    • S = { 0 errors, 1 errors , 2 errors , ... , inf }
  • ex3) three power plants, either generating electricity (1) or idle (0)
    
    • S = { (0,0,0), (0,0,1) , ... , (1,1,1) }
  • ex4) roll of die
    • S = { 1, 2, 3, 4, 5, 6 }
• Probabilites values : outcomes에 대한 확률
  • p1, p2, ... , pn
  • all of 0 <= pi <= 1
  • sum all of pi = 1
  • P(Oi) = pi
    • ex) software error probabilites
      • 0 error : 0.05, 1 error : 0.08, 2 error : 0.35, 3 error : 0.20, 4 err : 0.20, 5 err : 0.12
      • i err : 0 for i >= 6
      • there are at most 5 errors since the porbability values are zeroy for 6 or more errors
      • the most likely number of error is 2
      • 3 and 4 errors are equally likely in the software product
    • all of the possible outcomes is considered to be equally likely, we call it identical probability values
      • n outcomes -> 1/n probability
      • ex) fair die
        • S = { 1, 2, 3, 4, 5, 6 ), P = { 1/6, ... , 1/6 }
• Event : Event is subset of sample space S, The probability of event A, P(A)는 A의 원소의 확률을 다 더한 것
• Complements of Events (여집합) : P(A) + P(A`) = 1
  • elementary events or simple events : Events that consist of individual outcome
  • ex) event A = { 0 err, 1 err, 2 err }
    • P(A) = P(0 err) + P(1 err) + P(2 err) = 0.05 + 0.08 + 0.35 = 0.48
    • the probability of the complement of the A is 0.52
  • ex2) A = {2,4,6}  # one dice
    • P(A) = 1/2
  • ex3) A = the sum of scores of two dice is equal to 6
    • A = {(1,5), (2,4) , ... ,(5,1)}
    • P(A) = 5/36
    • B = at least one 6 recorded
    • P(B) = 11/36
    • P(B`) = 25/36
• Intersection = P(A and B)
  • 동시에 만족함
  • P(A and B) + P(A and B`) = P(A)
• mutually exclusive events : P(A and B) = 0
• Union = P(A or B)
  • P(A or B) = P(A and B`) + (A` and B) + P(A and B)
  • P(A or B) = P(A) + P(B) - P(A and B)
  • if A and B are mutually exclusive, then
  • P(A or B) = P(A) + P(B)
• Union Three Events : P(A or B or C) = P(A) + P(B) + P(C) - (P(A and B) + P(B and C) + P(A and C) + P(A and B and C)
  
  • if three events are mutually exclusive, then P(A or B or C) = P(A) + P(B) + P(C)
  • if all of the events are mutually exclusive events, then P(A1 or ... or An) = P(A1) + ... + P(An)
  • Sample Space partitions : A1, ... , An is mutually exclusive, so A1 or .. or An = S
    • each outcome in the sample space is within one and only one of the events Ai
• ! Conditional Probability : event A conditional on event B
  • P(A|B) = P(A and B) / P(B)
  • B의 환경 안에서 A가 발생할 확률
  • 만약 상호 배제라면 0, B가 A안에 속한다면 1, A가 B안에 속한다면 P(A)/P(B)
  • P(A|B) + P(A`|B) = 1 -> B의 환경 안에서 A가 발생한 확률 + B의 환경 안에서 A가 발생하지 않을 확률 = P(B) / P(B) = 1
• !! intersection of series of events Ai
  • P(A1 and ... and An) = P(A1) x P(A2 | A1) x P(A3 | A1 and A2) x ... x P(An | A1 and ... and An-1)
  • 각각의 부분이 합치는 곳에서 공통의 부분을 빼냄
  • P(A and B) = P(B) x P(A | B) = P(A) x P(B | A) -> A나 B의 관점이 아닌 S의 관점으로 바꿈
  • 식을 보면 B에 속한 A, 즉 A and B를 B 안의 관점에서 P(B)를 곱함으로써 관점을 S로 바꿈
  • P(A and B and C) = P(A and B) x P(C | A and B) : A와 B의 안에 속한 C를 A and B를 곱함으로써 전체에 대한 확률로 변경
  • so that is P(A)P(B|A)P(C|A and B) -> 시리즈로 나눈 것
• ! A and B are said to be independant events : 서로 영향 x != mutually exclusive
  • P(A|B) = P(A) : B가 일어나든말든 A의 확률에는 영향 x
  • P(A and B) = P(A) x P(B)
• independent evets A1, ... , An : P(A1 and ... and An) = P(A1) x P(A2) x ... x P(An)
  • !! ex) car warranties
    • plant 1 supplies 20%, 2 sup 24%, 3 sup 25%, 4 sub 31%
    • so probabilites is 0.20, 0.24, 0.25, 0.31
    • if P(claim|Plant 1) = 0.05, P(claim|Plant 2) = 0.11, P(claim|Plant 3) = 0.03, P(claim|Plant 4) = 0.08
      • 얘네들은 공장 1에서 만들었는데, 그 공장의 제품 중 0.05가 클레임 걸리는 확률임
    • so P(claim) is 0.05 x 0.20 + 0.11 x 0.24 + 0.03 +0.25 + 0.31 x 0.08
      • 이렇게 하면 전체 제품 중에 클레임 걸리는 비율을 알 수 있음
    • -> 조각난 claim을 하나로 합침
• !!! Law of Total Probability : A들에게 조각난 B의 조각들을 return해줌
  • P(B) = P(A1)P(B|A1) + ... P(An)P(B|An) 이때 모든 A는 상호 배제 상태
• !!!!!!! Bayes Theorem (세계관을 바꾸는 것)
  • P(Ai)와 P(B|Ai)를 이용해서 P(Ai|B)를 만들 수 있음
  • 앞에가 prior probabilities, 뒤에가 posterior probabilities임
  • P(Ai|B) = P(Ai)P(B|Ai) / sum all P(Aj)P(B|Aj)
  • 분모는 A에게 조각난 모든 B를 합함, 즉 P(B)가 됨
  • 분자는 특정 A에게 조각난 B의 조각임, 즉 P(Ai and B)가 됨. 
  • 따라서 기존에 A의 관점에서 보던 B를 B의 관점에서 A를 보게 됨
  • 위의 car warranties에서 P(Plant i|Claim)을 구하자
    • 즉 클레임이 걸렸을 때 각 공장의 확률을 구할 수 있음
    • P(Plant 1|Claim) = P(Claim|Plant 1) x P(Plant 1) / P(Claim)
    • = 0.05 x 0.20 / 0.0687 = 0.146임