SOLUTION: What is the probability of (AnBnC) if P(A) =.95; P(B) =.81; P(C) =.77

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Question 779525: What is the probability of (AnBnC) if P(A) =.95; P(B) =.81; P(C) =.77
Answer by stanbon(75887) About Me  (Show Source):
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What is the probability of (AnBnC) if P(A) =.95; P(B) =.81; P(C) =.77
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Assuming that events A,B, and C are independent,
P(A n B n C) = 0.95*0.81*0.77 = 0.5925
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Cheers,
Stan H.
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