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Probability P is a real valued function that assigns to eachevent E in the sample space S a number P(E)A probability of an event E is given by P(E) = N(E) / N(S)i.e. (No of outcomes in E) / (No of outcomes in S)
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ExampleExperiment : Throwing a coin twice Define events A and B as followsA = { At least one H is obtained } B= { at least one T isobtained} = { HT, TH, HH} = { TH, HT, TT}S= { HT,HH,TH,TT} Then N(S) = 4, N(A) = 3, N(B) = 3 a) P(A) = 3/4 b) P(A ∩ B) = 2/4 c) P(AC) = 1/4
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Axioms for Probability FunctionsAxiom 1 : For every event A in a sample space S, 1 ≥ P(A) ≥ 0Axiom 2 : P(S) = 1Axiom 3 : Let A and B be any two mutually exclusive eventsdefined over S. Then P(A U B) = P(A) + P(B)
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Example:Consider a random trial that can result in failure or success. Let 0 stand for failure, andlet 1 stand for success. Then we can consider the outcome space to be S = {0, 1}. Forany number p between 0 and 100%, define the function P as follows:P({1}) = p,P({0}) = 100% − p,P(S) = 100%,P({}) = 0.Then P is a probability distribution on S, as we can verify by checking that it satisfiesthe axioms:1. Because p is between 0 and 100%, so is 100% − p. The outcome space S has but four subsets: {}, {0}, {1}, and {0, 1}. The values assigned to them by P are 0, 1 − p, p, and 100%, respectively. All these numbers are at zero or larger, so P satisfies Axiom 1.2. By definition, P(S) = 100%, so P satisfies Axiom 2.3. The empty set and any other set are disjoint, and it is easy to see that P({}∪A) = P({}) + P(A) for any subset A of S. The only other pair of disjoint events in S is {0} and {1}. We can calculate P({0}∪{1}) = P(S) = 100% = (100% − p) + p = P({0}) + P({1}). Thus P satisfies Axiom 3.
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Theorem for Probability FunctionsTheorem 1If Ac is a complement of A , then P(Ac) = 1- P(A)Proof :S = Ac A,from Axiom 2 and 3, P(S) = 1 = P(A) + P(Ac) ,since A and Ac are ME P(Ac) = 1 – P(A)
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Theorem 2P(Ø) = 0, the impossible event has probabilityzero.Proof :Let S = Ac A, where A = ØThenS = Ac Ø Ac = S
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Theorem 3If A B, then P(A) ≤ P(B)Proof :We can write B as B = A (B A’) where A and(B A’) are ME.Thus, P(B) = P(A) + P(B A’)But P(B A’) 0 since A B This P(B) P(A)
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Theorem 4For any event A, P(A) ≤ 1Proof :We know that A S and P(S) = 1 P(A) ≤ P(S) = 1
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Theorem 5If A and B are any two events, then P(A B) = P(A) +P(B) - P(A B)Proof :A B=A ( Ac B)P(A B) = P(A) + P( Ac B) P( Ac B) = P(A B) - P(A) ……………….. (1)ButB = (A B) (Ac B)P(B) = P(A B) + P(Ac B) P( Ac B) = P(B) - P(A B) …………… (2) (1) = (2) P(A B) - P(A) = P(B) - P(A B) P(A B) = P(A) + P(B) - P(A B)
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Thm 6If A , B and C are any three events, thenP(A B) = P(A) +P(B) - P(A B)Proof :P(A) = P(A Bc) + P (A B) andP(B) = P(B Ac) + P (A B)adding these two equationP(A) + P(B) = [P(A Bc) + P(B Ac) + P (A B)] + P (A B) the sum in the bracket is P(A B).If we subtract P (A B) both sides of the equation,the result follws.
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