Good day everyone,

I am currently looking at a self-help excercise which goes about exhibiting the following behavior which really puzzled me. $$> P(D) = P(A \cap D) + P(B \cap D) + P(C \cap D)$$ After drawing out the Venn diagram, it just doesnt seem to make sense.

The original question that prompted my question as follows:

• +1, but technically 'that's true only if' is false: it suffices that the intersections are of probability 0 and that the universe minus $A\cap B\cap C$ is of probability 0. In addition, the statement could be true 'by accident', e.g., let $P(D)=1,~A\subset D,~ P(A)=1/3,~A=B=C$. Suggestion: edit to remove 'only' from the beginning sentence. Commented May 5, 2014 at 12:42
• "and if the union of those three sets is equal to the universe set." This part can't be quite right? I can certainly draw a Venn diagram where the first clause of your sentence is true, and the pre-fixed condition is true (imagine A, B an C flush, but not overlapping, D intersects all three, but nothing else, i.e if P$(D \cap (A \cup B \cup C)^{c})$ = 0 is also true. Commented May 5, 2014 at 12:57