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Theory of Probability UNIT - I Probability Theory Probability Distributions
Contents ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Introduction ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
Simple Definitions ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],Possible solutions – Ace to King Exhaustive no. of cases – 52 In drawing a card from a well shuffled standard pack of playing cards Possible solutions – 1,2,3,4,5,6 Exhaustive no. of cases – 6 In a throw of an unbiased cubic die Possible solutions – Head/ Tail Exhaustive no. of cases – 2 In a tossing of an unbiased coin Collectively Exhaustive Events Experiment
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],Card is a spade or heart Card is a diamond or club Card is a king or a queen In drawing a card from a well shuffled standard pack of playing cards Occurrence of 1 or 2 or 3 or 4 or 5 or 6 In a throw of an unbiased cubic die Head/ Tail In a tossing of an unbiased coin Mutually Exclusive Events Experiment
[object Object],[object Object],[object Object],Any card out of 52 is likely to come up In drawing a card from a well shuffled standard pack of playing cards Any number out of 1,2,3,4,5,6 is likely to come up In a throw of an unbiased cubic die Head is likely to come up as a Tail In a tossing of an unbiased coin Collectively Exhaustive Events Experiment
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Types of Probability ,[object Object],[object Object],[object Object],[object Object]
Mathematical/ Classical/ ‘a priori’ Probability ,[object Object],[object Object],[object Object],[object Object],Number of favorable cases  m p = P(E) = ------------------------------ = ---- Total number of equally likely cases  n
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],Limitations of Classical definition
[object Object],[object Object],[object Object],[object Object]
Relative/ Statistical/ Empirical Probability ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],m p = P(E) = Lt  ---- N ->     N
[object Object],[object Object],[object Object],Limitations of Statistical/ Empirical method
The Axiomatic Approach ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],n  n P ( U A i ) = ∑  P( A i  ) i=1   i=1
The Objective and Subjective Approach ,[object Object],[object Object],[object Object],[object Object],[object Object]
Theorems of Probability ,[object Object],[object Object],[object Object]
Addition theorem when events are Mutually Exclusive ,[object Object],[object Object],[object Object],P(A or B) or P(A U B) = P(A) + P(B) P(A or B or C) = P(A) + P(B) + P(C)
Addition theorem when events are not Mutually Exclusive (Overlapping or Intersection Events) ,[object Object],[object Object],P(A or B) or P(A U B) = P(A) + P(B) – P(A ∩ B)
Multiplication theorem ,[object Object],[object Object],[object Object],[object Object],P(A and B) or P(A ∩ B) = P(A) x P(B) P(A, B and C) or P(A ∩ B ∩ C) = P(A) x P(B) x P(C)
How to calculate probability in case of Dependent Events P(A U B) = P(A) + P(B) P(A U B) = P(A) + P(B) – P(A ∩ B) P(A ∩ B) = P(A) + P(B) – P(A U B) P(A ∩ B) = P(A) - P(A ∩ B) P(A ∩ B) = P(B) - P(A ∩ B) P(A ∩ B) = 1 - P(A U B) P(A U B) = 1 - P(A ∩ B)  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Formula Case
How to calculate probability in case of Independent Events P(A ∩ B) = P(A) x P(B) P(A ∩ B) = P(A) x P(B) P(A ∩ B) = P(A) x P(B) P(A ∩ B) = P(A) x P(B) P(A U B) = 1 - P(A ∩ B) = 1 – [P(A) x P(B)] P(A U B) = 1 - P(A ∩ B)  = 1 – [P(A) x P(B)] P(A ∩ B) + P(A ∩ B) = [P(A) x P(B)] +  [P(A) x P(B)]  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Formula Case
Problem ,[object Object],[object Object],[object Object],[object Object],[object Object]
Solution ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Probability Rules
Probabilities under conditions of Statistical Independence ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example: Marginal Probability - Statistical Independence ,[object Object],[object Object],[object Object],[object Object],Marginal Probability of an Event P(A) = P(A)
[object Object],[object Object],[object Object],[object Object],[object Object],Example: Joint Probability - Statistical Independence Joint Probability of 2 Independent Events P(AB) = P(A) * P(B)
[object Object],0.125 T 1 T 2 T 3 0.125 T 1 T 2 H 3 0.125 T 1 H 2 T 3 0.125 T 1 H 2 H 3 0.125 H 1 T 2 T 3 0.25 T 1 T 2 0.125 H 1 T 2 H 3 0.25 T 1 H 2 0.125 H 1 H 2 T 3 0.25 H 1 T 2 0.5 T 1 0.125 H 1 H 2 H 3 0.25 H 1 H 2 0.5 H 1 3 Toss 2 Toss 1 Toss
[object Object],[object Object],[object Object],[object Object],Example: Conditional Probability - Statistical Independence Conditional Probability for 2 Independent Events P(B|A) = P(B)
Probabilities under conditions of Statistical Dependence ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example ,[object Object],[object Object],[object Object],[object Object],[object Object],0.1 10 0.1 9 0.1 8 Gray & Striped 0.1 7 0.1 6 Gray & Dotted 0.1 5 Colored & Striped 0.1 4 0.1 3 0.1 2 Colored & Dotted 0.1 1 Probability of Event Event
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Marginal Probability - Statistically Dependent
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Example: Joint Probability - Statistically Dependent Joint probability for Statistically Dependent Events P(BA) = P(B|A) * P(A)
[object Object],[object Object],[object Object],Example: Conditional Probability - Statistically Dependent Conditional probability for Statistically Dependent Events   P(BA) P(B|A) =  ---------- P(A)
We know that there are 4 colored balls, 3 of which are dotted & one of it striped.     P(DC)  0.3 P(D|C) = --------- = ------      P(C)  0.4   = 0.75 P(DC) = Probability of colored & dotted balls  (3 out of 10 --- 3/10) P(C) = 4 out of 10 --- 4/10
Revising Prior Estimates of Probabilities: Bayes’ Theorem ,[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
Bayes Theorem ,[object Object],[object Object],P(A | B i ). P(B i ) P(B i  | A) = ---------------------- ∑  P(A | B i ). P(B i )
[object Object],[object Object],[object Object],[object Object]
Problem ,[object Object],[object Object],[object Object]
Solution ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Glossary of terms ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Fundamentals Probability 08072009

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  • 42. We know that there are 4 colored balls, 3 of which are dotted & one of it striped. P(DC) 0.3 P(D|C) = --------- = ------ P(C) 0.4 = 0.75 P(DC) = Probability of colored & dotted balls (3 out of 10 --- 3/10) P(C) = 4 out of 10 --- 4/10
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