Question
Asked Sep 13, 2019
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Let EF and G be three events in S with P(E) = 0.4, P(F) = 0.5, P(G) = 0.29, P(E ∩ F) = 0.28, P(E ∩ G) = 0.11, P(F ∩ G) = 0.14, and P(E ∩ F ∩ G) = 0.06. Find P(EC ∪ FC ∪ GC).

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Expert Answer

Step 1

Given Data

P(E) = 0.4, P(F) = 0.5,P(G) = 0.29, P (E ∩ F) = 0.28, P(E ∩ G) = 0.11, P(F ∩ G) = 0.14, and P(E ∩ F ∩ G) = 0.06.

De-Morgan Law states that

Let E,F and G be three events

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EC U FC U GC(ENFNG)C

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Step 2

The required probability is ...

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P(EC U FC U G)= P(ENFNG) P(EC U FC U G = 1 - P(E n FnG) 1 - 0.06 = 0.94

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