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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.25 and P(B | A2) = 0.05. (a) Are A1 and A2 mutually exclusive? They mutually exclusive. How could you tell whether or not they are mutually exclusive? P(B | A1) ≠ P(B | A2) P(A1) + P(A2) = 1 P(A1 ∩ A2) = 0 P(A1) ≠ P(A1 | A2) P(A2) ≠ P(A2 | A1) (b) Compute P(A1 ∩ B) and P(A2 ∩ B). P(A1 ∩ B) = P(A2 ∩ B) = (c) Compute P(B). (d) Apply Bayes' theorem to compute P(A1 | B) and P(A2 | B). (Round your answers to four decimal places.) P(A1 | B) = P(A2 | B) =
The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.25 and P(B | A2) = 0.05. (a) Are A1 and A2 mutually exclusive? They mutually exclusive. How could you tell whether or not they are mutually exclusive? P(B | A1) ≠ P(B | A2) P(A1) + P(A2) = 1 P(A1 ∩ A2) = 0 P(A1) ≠ P(A1 | A2) P(A2) ≠ P(A2 | A1) (b) Compute P(A1 ∩ B) and P(A2 ∩ B). P(A1 ∩ B) = P(A2 ∩ B) = (c) Compute P(B). (d) Apply Bayes' theorem to compute P(A1 | B) and P(A2 | B). (Round your answers to four decimal places.) P(A1 | B) = P(A2 | B) =
Answer:(a)andare indeed mutually-exclusive.(b), whereas.(c).(d), whereasStep-by-step explanation:(a)means that it is impossible for eventsandto happen at the same time. Therefore, eventandaremutually-exclusive.(b)By the definition ofconditional probability:.Rearrange to obtain:.Similarly:.(c)Note that:.In other words,andarecollectively-exhaustive. Sinceandare collectively-exhaustive and mutually-exclusive at the same time:.(d)By Bayes' Theorem:.Similarly:....
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