1.If P(A) = 0.45, P(B) = 0.8, P(AUB) = 0.72, find the probability of A and B.?
2016-08-09 4:53 am
2. If P(A) = 0.3, P(B) = 0.55, find the probability of A or B if we know that A and B are mutually exclusive.
回答 (1)
2016-08-09 5:04 am
OK, it helps if you draw a Venn diagram.
1) P(A U B) = P(A) + P(B) - P (A ∩ B)
Rearranging gives :
P (A ∩ B) = P(A) + P(B) - P (A U B)
P (A ∩ B) = 0.45 + 0.8 - 0.72
P (A ∩ B) = 0.53
2) Note that if two events are mutually exclusive, then they have nothing in common and thus the intersection is 0.
So P (A ∩ B) = 0
P (A U B) = 0.3 + 0.55 + 0
P (A U B) = 0.85
Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!
1) P(A U B) = P(A) + P(B) - P (A ∩ B)
Rearranging gives :
P (A ∩ B) = P(A) + P(B) - P (A U B)
P (A ∩ B) = 0.45 + 0.8 - 0.72
P (A ∩ B) = 0.53
2) Note that if two events are mutually exclusive, then they have nothing in common and thus the intersection is 0.
So P (A ∩ B) = 0
P (A U B) = 0.3 + 0.55 + 0
P (A U B) = 0.85
Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!
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