Question 1203560: Let P(A) = 1/2, P(B) 1/8, and P(C) = 1/4, where A, B, and C are mutually exclusive.
Find the following:
A) P(A u B u C)
B) P(A' n B' n C')
Answer by math_tutor2020(3817)   (Show Source):
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Problem 1
A,B,C are mutually exclusive means
P(A n B) = 0
P(A n C) = 0
P(B n C) = 0
P(A n B n C) = 0
where the "n" refers to set intersection.
Something like P(A n B) = 0 means events A and B cannot happen simultaneously. They are disjoint events.
Use the inclusion-exclusion principle to say the following:
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(B n C) - P(A n C) + P(A n B n C)
P(A u B u C) = (1/2) + (1/8) + (1/4) - 0 - 0 - 0 + 0
P(A u B u C) = (1/2) + (1/8) + (1/4)
P(A u B u C) = (4/8) + (1/8) + (2/8)
P(A u B u C) = (4+1+2)/8
P(A u B u C) = 7/8
Or as a shortcut, because events A,B,C are mutually exclusive we can use this formula
P(A u B u C) = P(A) + P(B) + P(C)
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Problem 2
P(A' n B' n C') = P( (A' n B') n C')
P(A' n B' n C') = P( (A u B)' n C') ... De Morgans Law
P(A' n B' n C') = P( ((A u B) u C)') ... De Morgans Law
P(A' n B' n C') = P( (A u B u C)')
P(A' n B' n C') = 1 - P(A u B u C) ... complement law
P(A' n B' n C') = 1 - (7/8) .... use result of problem 1
P(A' n B' n C') = (8/8) - (7/8)
P(A' n B' n C') = (8-7)/8
P(A' n B' n C') = 1/8
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