P(aub)=p(a)+p(b)-p(ab)=0+0-0=0 done.
Hey people. Im sitting here with an assignment.
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It is given that P(A) = P(B) = 0
Then i have to prove that P(A U B) = 0.
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We are supposed to use the theorem 3.1:
Let P be a probability measure. Then we have that
P(A\B) = P(A) - P(B) if A contains B
and
P(Ac) = 1 - P(A) where Ac is A complement.
furthermore, P is increasing in the sense that
if A contains B it implies that P(A) is greater than or equal to P(B)
and P is continuous in the sense that An (arrow up) A => P(An) (arrow up) P(A) and the same the other way around.
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I hope it makes sense and i hope someone can help me.
Its a fairly simple thing, but somehow i cant wrap my head around it.
what does this mean.............
'As far as i know P(AB) isnt equal to P(AB)?'
so , very interesting, but..............
AB means intersection and AB is contained in A , also B
So it has probability zero
anyhow you still would have
P(AUB)=-P(AB)which is less than or eqal to zero...... giving you 0=0.
It's 3am and I don't have time for this.
yes, it's just notation
if you prefer
SEE (1) on http://www.weibull.com/SystemRelWeb/...ity_theory.htm