If the probability of an event $A$ is $1$, and $B$ is any event, then $A$ and $B$ are independent.

Question

If the probability of an event $A$ is $1$, and $B$ is any event, then $A$ and $B$ are independent.

This seems intuitively true. If we know for sure that $A$ happens, this does not give information about $B$.

However, I am not able to prove it, although I think it must be quite easy to prove.

So given is that $P(A) =1$, and now we must prove that $P(B) = P(A)P(B) = P(A \cap B)$ and it suffices to show that $P(A \cap B) \geq P(B)$, since $A \cap B \subseteq B$

Answer 1

$$P(A)P(B)=1 \times P(B)=P(B)=P(A\cap B)+P(A^c \cap B) = P(A\cap B)+0=P(A\cap B)$$

Answer 2

If $P(A)=1$, then $A$ is the sample space. Hence: $$P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B)}{1}=P(B).$$

Answer 3

The mathematics is straight forward but intuitively any event happening with probability $1$ is independent from any other event despite the characteristics of that.

Answer 4

Assume $A,B$ are events such that $P(A) = 1$. Consider $P(A \cap B)$. Since $P(A) = 1 = P(S)$, where $S$ is the sample space, we have $A = S$. As such, $B \subset S$ and so $P(A\cap B) = P(B) = 1\cdot P(B) = P(A) \cdot P(B)$. Thus, $A,B$ are independent events.