Data Science Classroom Series – 02/Nov/2021

Independence of Events in Probability

  • Events are often classified as
    • Dependent Events:
      • General Examples
        • If you park vehicle in unauthorized place, you are more likely to get a parking ticket
        • You must buy a lottery ticket to have a chance of winning
    • Independent Events
      • General Events:
        • Winning a Card game and running out of stock at home
        • Taking an rapido ride and getting a gift from friend
  • Two Events A and B are considered independent if any of the following is true
    • P(A|B) = P(A)
    • P(B|A) = P(B)
    • P(A ∩ B) = P(A)*P(B)
  • If none of the above holds good we call it dependent
  • Example-1: Consider the following events
    • A: odd number is rolled
    • B: even number is rolled
    • C: {1,2}
    • Are A and B independent?
    P(A) =  3/6 = 1/2
    P(B) =  3/6 = 1/2
    P(A|B) = P (A ∩ B)/ P(B)
    P (A ∩ B) = 0
    p(A|B) = 0
    # Since none of the above cases are true 
    A and B are dependent events
    
    • Are A and C independent?
    P(A) = 1/2
    P(C) = 2/6 = 1/3
    P(A|C) = 1/2 = P(A)
    Events are independent
    
  • Example-2: Three Car brands X,Y and Z are ranked by judges
    • A – Car X is ranked higher than Y
    • B – Car X is ranked highest
    • C – Car X is ranked second best
    • D – Car X is ranked worst
Sample Space 
E1: XYZ  E2: XZY E3: YXZ
E4: YZX  E5: ZXY E6: ZYX

A = { E1, E2, E5 }
B = { E1, E2 }
C = { E3, E5 }
D = { E4, E6}

P(A) = 3/6 = 1/2
P(A|B) = P(A ∩ B) / P(B) = 2/6 / 2/6 = 1
P(A|C) = P(A ∩ C)/P(C) = 1/6/2/6 = 1/2
P(A|D) = P(A ∩ D)/P(D) = 0

  • Result : A and C are independent & A,B and A, D are dependent

Multiplicative law of Probability

  • The probability of intersection of two events A and B is
P(A ∩ B )= P(B)*P(A|B)
  • If the events are independent “`P(A ∩ B) = P(A) * P(B)
  • Example: Lets consider the Dependent events, when you roll a die
    • A: Odd Number
    • B: number < 4
P(A) = 3/6 = 1/2
P(B) = 3/6 = 1/2
P(A|B)= 2/3
P(B|A) = 2/3

P(A ∩ B) = 1/2 * 2/3 = 1/3

Additive Law of Probability

  • Probability of union of events P(A U B) = P(A) + P(B) - P(A ∩ B)
  • If A and B are mutually exclusive P(A U B) = P(A) + P(B)

Law of Total Probability

  • Partitions: If S is the Sample space of the experiment and given k disjoint events B1, B2….BK Preview Preview
  • we can say that these events form the Partition of S
  • Law of Total Probability:
    • Suppose B1….Bk are the partitions over the space S and P(Bj) > 0 for j= 1…k
    • Then for every A in S Preview
    • We can have a conditional version Preview
  • Example:
    • A player plays a game where the final score is somewhere from 1 to 50 (equally likely)
    • Players score after first game is X
    • Player continues to play until they obtain another score Y such that Y>X
    • What is P(Y=50)
  • Lets define B i = P(X=i), the value is likely to be any number i, i+1 …. , 50 and Each of these (51-i) values for Y are equally likely Preview

Bayes Theorem

  • This describes probability of an event bas on prior knowledge of conditions that might be releate to the event
           P(B|A) P(A)
P(A|B) = ------------
             P(B)
  • There is an extended form of Bayes theorem for partitions
  • Let events B1,….., B k form a partition set over S such that P(B j) > 0 for j = 1…k abd let A be the event such that P(A)>0 Then for i = 1..K Preview
  • Conditional version Preview
  • Example:
    • You can test a person for the presence or absence of particular disease
    • The test is 90% reliable
    • If the Person has the disease P(test positive) = 0.9
    • If the person does not have the disease P(test positive) = 0.1
    • Chance of having this disease is 1 in 10000
    • You have tested positive. What is the probability that you actually have a disease
  • Solution:
    • Let B1 denote you have the disease and B2 denote that you do not
    • B1 and B2 will for partition
    • Let A denote the event that the response to test is positive Preview

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.

About learningthoughtsadmin