- #1
FrankJ777
- 140
- 6
I was trying to solve a problem about calculating the probability of failures of a communications link with several nodes, and figured I need to use Demorgan's law since it would involve non mutually exclusive events. So I tried to go back to the basics to make sure I understand the axioms of probability and prove to my self that this is correct using Venn diagrams, but I can't make the numbers come out right trying to apply Demorgan's Law. Here is the example:
P(A) = .3
P(B) = .2
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = .3 + .2 - .05 = .45
But if I try to apply Demorgans's I don't get what I expect.
P(A∪B)' = P(A' ∩ B')
P(A∪B)' = (1-P(A))(1-P(B))
P(A∪B)' = (1-.3)(1-.2)
P(A∪B)' = (.7)(.8) = .56
but clearly P(A∪B)' = .55
so where am i going wrong? All parts of my Venn diagram add up to 1. Also I believe that the identity P(A∩B) = P(A)P(B) but this doesn't seem to be correct either as .3 x .2 ≠ .05. So I guess I'm misapplying the use of the Venn diagram but I'm not sure how. Can someone please explain what I'm doing wrong?
Thanks in advance.
P(A) = .3
P(B) = .2
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = .3 + .2 - .05 = .45
But if I try to apply Demorgans's I don't get what I expect.
P(A∪B)' = P(A' ∩ B')
P(A∪B)' = (1-P(A))(1-P(B))
P(A∪B)' = (1-.3)(1-.2)
P(A∪B)' = (.7)(.8) = .56
but clearly P(A∪B)' = .55
so where am i going wrong? All parts of my Venn diagram add up to 1. Also I believe that the identity P(A∩B) = P(A)P(B) but this doesn't seem to be correct either as .3 x .2 ≠ .05. So I guess I'm misapplying the use of the Venn diagram but I'm not sure how. Can someone please explain what I'm doing wrong?
Thanks in advance.