Probability

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Conditional probability is the probability of event E given that event F has already occurred. Formula: P(E|F) = P(E∩F)/P(F), where P(F) ≠ 0. This measures how the occurrence of F affects the likeliho

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Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} F = {THH, THT, TTH, TTT} E∩F = {THH} P(E|F) = Number of favorable outcomes in F / Total outcomes in F = 1/4 Alternatively: P(E|F) = P(E∩F)/P(F

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Property 1: P(S|F) = P(F|F) = 1 Property 2: P((A∪B)|F) = P(A|F) + P(B|F) - P((A∩B)|F) For disjoint events: P((A∪B)|F) = P(A|F) + P(B|F) Property 3: P(E'|F) = 1 - P(E|F)

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For two events E and F: P(E∩F) = P(E)·P(F|E) = P(F)·P(E|F) This theorem helps find the probability of simultaneous occurrence of two events. It's valid when P(E) ≠ 0 and P(F) ≠ 0.

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Let E₁ = first ball is red, E₂ = second ball is red P(E₁) = 4/10 = 2/5 P(E₂|E₁) = 3/9 = 1/3 (after removing one red ball) Using multiplication theorem: P(both red) = P(E₁∩E₂) = P(E₁)·P(E₂|E₁) = (2/5)·

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Two events E and F are independent if the occurrence of one does not affect the probability of the other. Mathematical conditions: 1. P(E|F) = P(E) (provided P(F) ≠ 0) 2. P(F|E) = P(F) (provided P(E)

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P(E) = 13/52 = 1/4 P(F) = 4/52 = 1/13 P(E∩F) = P(ace of spades) = 1/52 Check: P(E)·P(F) = (1/4)·(1/13) = 1/52 Since P(E∩F) = P(E)·P(F), events E and F are independent.

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Independent Events: - P(E∩F) = P(E)·P(F) - Can have common outcomes - Occurrence of one doesn't affect the other Mutually Exclusive Events: - E∩F = ϕ (no common outcomes) - P(E∩F) = 0 - Cannot occur