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Questions › JNU › JNU SIS 2018 › Q18
2 marks, −1Statistics › Probability and Bayes' theorem › Addition rule, complements and inclusion-exclusionModerateabout 2 min

JNU SIS 2018 Question 18: Probability and Bayes' theorem

AAA, BBB, CCC are independent events such that P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.6P(B) = 0.6P(B)=0.6, P(C)=0.1P(C) = 0.1P(C)=0.1. Then P(Ac∩Bc∪C)P(A^c \cap B^c \cup C)P(Ac∩Bc∪C) (where AcA^cAc and BcB^cBc are the complements of events AAA and BBB, respectively, in the sample space) is

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A, B, C are independent events such that P(A) = 0.5, P(B) = 0.6, P(C) = 0.1. Then P(A^c ∩ B^c ∪ C) (where A^c and B^c are the complements of events A and B, respectively, in the sample space) is (A) 0.69 (B) 0.73 (C) 0.71 (D) 1

Also written as: A, B, C are independent events such that P(A) = 0.5, P(B) = 0.6, P(C) = 0.1. Then P(A^c intersection B^c union C) (where A^c and B^c are the complements of events A and B, respectively, in the sample space) is

Probability and Bayes' theorem has come up 190 times in 7 exams since 2004.

What this question tests

This question is on Addition rule, complements and inclusion-exclusion, part of Probability and Bayes' theorem in Statistics.

The probability of at least one of several events, and of none.

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Question from the JNU SIS 2018 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.