Explanation Formula used: Distributive law: A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) Probability: If each simple event in sample space S is likely to occur as any other, then the probability of an arbitrary event E is given by, P ( E ) = Number of elements of E Number of elemenst of S = n ( E ) n ( S ) Complement of an event: Suppose the sample space S = { e 1 , ... , e n } is divided into two subsets E and E ′ such that E ∩ E ′ = ∅ that is E and E ′ are mutually exclusive and E ∪ E ′ = S , then E ′ is called the complement of E relative to S and is defined as, P ( E ) = 1 − P ( E ′ ) or P ( E ′ ) = 1 − P ( E ) Conditional probability: Let A and B be two events then the conditional probability is given by, P ( A | B ) = P ( A ∩ B ) P ( B ) or P ( B | A ) = P ( A ∩ B ) P ( A ) Proof: Let A and B be two events, then A ′ is the complement events for A . Since the events A and A ′ are complement events, then A ∪ A ′ = U , where U is the universal set. Consider, B = U ∩ B . B = U ∩ B = ( A ∪ A ′ ) ∩ B = ( A ∩ B ) ∪ ( A ′ ∩ B ) [ Using distributive law ] Also, it is known that ( A ∩ B ) ∩ ( A ′ ∩ B ) = ∅ because there would be no common elements for the two sets

College Mathematics for Business, Economics, Life... (Instructor's)

14th Edition

ISBN: 9780134676104

Author: Barnett

Publisher: PEARSON

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Chapter 8.3, Problem 78E

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To show: That P(A|B)+P(A′|B)=1.

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Chapter 8.3, Problem 77E

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Chapter 8 Solutions

College Mathematics for Business, Economics, Life... (Instructor's)

Ch. 8.1 - Repeat Example 1 for The outcome is a number...Ch. 8.1 - Prob. 2MP Ch. 8.1 - Prob. 3MP Ch. 8.1 - Suppose in Example 4 that after flipping the...Ch. 8.1 - Prob. 5MP Ch. 8.1 - Prob. 6MP Ch. 8.1 - Prob. 7MP Ch. 8.1 - Prob. 8MP Ch. 8.1 - Prob. 1ED Ch. 8.1 - Prob. 1E

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