Explanation Define the events: A = { Children 1 and 2 have the same gene } B = { Children 1 and 3 have the same gene } C = { Children 2 and 3 have the same gene } Consider that A ∩ B = B ∩ C = A ∩ C = A ∩ B ∩ C . Here, three children have the same gene. Therefore, P ( A ∩ B ) = P ( B ∩ C ) = P ( A ∩ C ) = P ( A ∩ B ∩ C ) . In this situation, it is enough to obtain P ( A ∩ B ) . The total number of possible combinations is 2 × 2 × 2 = 8 because each children have any of the two possible genes. All three children have either “ A gene” or “ a gene” when three children have the same gene. In a total of eight combinations, there are two possible combinations for the event of ( A ∩ B ) . Therefore, the probability of ( A ∩ B ) is, P ( A ∩ B ) = 2 8 = 1 4 Hence, P ( A ∩ B ) = P ( B ∩ C ) = P ( A ∩ C ) = P ( A ∩ B ∩ C ) = 1 4 . Assume that P ( A ) = P ( B ) = P ( C ) . In this situation, it is enough to obtain P ( A ) because each event indicates that at least two out of three children have the same gene. The number of possible combinations for event A is 2 × 1 × 2 = 4 . Therefore, the probability of event A is, P ( A ) = 4 8 = 1 2 Hence, P ( A ) = P ( B ) = P ( C ) = 1 2

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ISBN: 9780100244702

Author: RICE

Publisher: YUZU

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Chapter 1.8, Problem 80P

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Show that the given events are pairwise independent but not mutually independent.

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Chapter 1 Solutions

EBK MATHEMATICAL STATISTICS AND DATA AN

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Show that the given events are pairwise independent but not mutually independent.

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Show that the given events are pairwise independent but not mutually independent.

Chapter 1 Solutions