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Questions › DSE › DSE 2014 › Q15
1 mark, −0.333333333333333Statistics › Probability and Bayes' theorem › IndependenceHardabout 3 min

DSE 2014 Question 15: Probability and Bayes' theorem

Let events E,F,GE,F,GE,F,G be pairwise independent with Pr⁡(G)>0\Pr(G)>0Pr(G)>0 and Pr⁡(E∩F∩G)=0\Pr(E\cap F\cap G)=0Pr(E∩F∩G)=0. Let XCX^CXC denote the complement of XXX. Then Pr⁡(EC∩FC∣G)=\Pr(E^C\cap F^C\mid G)=Pr(EC∩FC∣G)=

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Let events E,F,G be pairwise independent with Pr (G)>0 and Pr (E∩ F∩ G)=0. Let X^C denote the complement of X. Then Pr (E^C∩ F^C| G)= (A) Pr (E^C)+Pr (F^C) (B) Pr (E^C)-Pr (F^C) (C) Pr (E^C)-Pr (F) (D) None of the above

Also written as: Let events E,F,G be pairwise independent with Pr (G)>0 and Pr (E intersection F intersection G)=0. Let X^C denote the complement of X. Then Pr (E^C intersection F^C| G)=

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

What this question tests

This question is on Independence, part of Probability and Bayes' theorem in Statistics.

When one event tells you nothing about another, and how it differs from being mutually exclusive.

TestStop has 29 past-paper questions on this subtopic.

Practise Probability and Bayes' theoremMore on Independence

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Question from the DSE 2014 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.