Proof that P(A) = 1, P(B) = 1, then P(AB) = 1

[hholzer]

Theorem:
If P(A) = 1, P(B) = 1, then P(AB) = 1

My book starts out with the proof as follows:

P(A U B) >= P(A) = 1, so P(A U B) = 1

How do they reach such a conclusion?

Things I know:
P(A U B) = P(A) + P(B) - P(AB)

How can I use that to be sure that P(A U B) = 1?

 

[statdad]

How does the fact that $$A \subset A \cup B$$

verify that $$P(A \cup B) \ge P(A)$$?

Second: IF you understand why the first comment is true, how does it, coupled with $$P(A) = 1$$ show $$P(A \cup B) = 1$$?

 

[gb7nash]

Theorem:
If P(A) = 1, P(B) = 1, then P(AB) = 1

My book starts out with the proof as follows:

P(A U B) >= P(A) = 1, so P(A U B) = 1

It is true that P(A U B) >= P(A). We know that P(A) = 1. Read P(A U B) as the probability of A OR B occurring (maybe both). Well, the probablility of A is 1, so A U B will always be true regardless of if B happens or not. So since P(A U B) >= P(A), P(A) = 1 and a probablility greater than 1 is not possible, the only value we can have for P(A U B) is 1. So:

P(A U B) = P(A) + P(B) - P(AB) -> 1 = 1 + 1 - P(AB) -> P(AB) = 1