2006-12-03 14:15:06 · 6 answers · asked by oking 1 in Science & Mathematics ➔ Mathematics
In general Pr(A U B U C) = Pr(A) + Pr(B) +Pr(C) – Pr(A∩B) –Pr(A∩C) – Pr(B∩C) +Pr(A∩B∩C).
But Pr(A∩B), Pr(A∩C), Pr(B∩C), Pr(A∩B∩C) all equal 0, therefore
Pr(A U B U C) = Pr(A) + Pr(B) +Pr(C)
2006-12-03 14:26:25 · answer #1 · answered by Dennis T 1 · 1⤊ 0⤋
If they're mutually exclusive, it's impossible for any two of them to take place, let alone all three, so 0.
2006-12-03 22:17:41 · answer #2 · answered by Amy F 5 · 0⤊ 1⤋
i have know idea what you are trying to say but just take the number of possibilties for each one and times it by 3 then take that number and put it under one and turn that fraction into a percent
2006-12-03 22:18:35 · answer #3 · answered by bill f 3 · 0⤊ 0⤋
The probability for the union of A, B, and C:
P(A) + P(B) + P(C) - P(A*B*C)
2006-12-03 22:19:51 · answer #4 · answered by AibohphobiA 4 · 0⤊ 2⤋
P(A U B U C)=P(A)+P(B)+P(C)
2006-12-03 22:16:43 · answer #5 · answered by Smitty Carmichael 2 · 2⤊ 0⤋
P(A or B or C) = P(A) + P(B) + P(C)
2006-12-03 22:25:39 · answer #6 · answered by Biznachos 4 · 0⤊ 0⤋