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

  • Thread starter hholzer
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
statdad
Homework Helper
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How does the fact that [tex] A \subset A \cup B [/tex]

verify that [tex] P(A \cup B) \ge P(A) [/tex]?

Second: IF you understand why the first comment is true, how does it, coupled with [tex] P(A) = 1 [/tex] show [tex] P(A \cup B) = 1 [/tex]?
 
  • #3
gb7nash
Homework Helper
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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 occuring (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
 

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