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Exercise
ABy Admin
Apr 29'23
Answer
Solution: D
Let B, C, and D be the events of an accident occurring in 2014, 2013, and 2012, respectively. Let A = B ∪ C ∪ D .
[[math]]
\operatorname{P}[B | A] = \frac{P[ A | B]P[ B]}{P[ A | B][ P[ B] + P[ A | C ]P[C ] + P[ A | D]P[ D]}
[[/math]]
Use Bayes’ Theorem
[[math]]
\frac{(0.05)(0.16)}{(0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20)} = 0.45.
[[/math]]