IF A=B with probability 1 then P(A) = P(B)=P(AB)

should this be solved as using addition law as follows ...

P (A ∪ B) = P (A) + P (B) = 1

i am not sure ... does anyone have any input?

Printable View

- February 7th 2009, 11:35 PMjeffProbability Proof Question
IF A=B with probability 1 then P(A) = P(B)=P(AB)

should this be solved as using addition law as follows ...

P (A ∪ B) = P (A) + P (B) = 1

i am not sure ... does anyone have any input? - February 9th 2009, 01:00 PMdely84
This is how I see it:

P(AB)=P(A¬B).P(B) but P(A¬B)=1 as A=B with probability one

=> P(AB)=P(B)

Analogously, P(AB)=P(B¬A)P(A) => P(AB)=P(A)

"¬" denotes conditional probability(sorry don't know how to use latex)