Can we identify if 2 events $A$ and $B$ are independent (or dependent) without knowing $P(A \cap B)$?

The above diagram certainly tells us that that the 2 events $$A$$ and $$B$$ are not mutually exclusive. But what do we know for sure about their independence. Can we tell anything about their independence without knowing the values for $$P(A \cap B)$$ .

Let $$P(A) = 0.4$$ and $$P(B) = 0.8$$

• You can obtain upper and lower bounds on $P(A \cap B)$; the independence case sits not too far from the middle of those bounds. Commented Oct 17, 2021 at 9:30
• Can you give me the upper and lower bounds for this particular example ? And also, the venn diagrams tell us nothing about independence correct ? Only knowing what $A$ and $B$ , or the value of $P(A∩B)$ can tell us anything about independence ? These are the main questions I have Commented Oct 17, 2021 at 9:41
• Note that P(A) + P(B) = 1.2. Clearly you can't have a probability > 1. How large is the smallest allowed P(A∩B) to keep P(A∪B) a valid probability? By the same token, what's the largest possible P(A∩B)? Commented Oct 17, 2021 at 11:01

I think you can identify two variables are independent instead by checking if

$$P(A|B) = P(A)$$

For all possible values of $$A$$ and $$B$$

• What is the definition of $P(A\mid B)$? and how can anyone check whether or not $P(A\mid B)$ equals $P(A)$ or not without first determining the value of $P(A\cap B)$ as a first step in calculating the value of $P(A\mid B)$? Commented Oct 16, 2021 at 20:22

I think you cannot tell something about two events $$A, B$$ independence if you just know the marginal probabilities $$P(A), P(B)$$. If you also had information in the probability of union $$P(A\cup B)$$ then you might could say something with the formula $$P(A\cup B)= P(A) + P(B) - P(A\cap B)$$. Otherwise, there are two unknown quantities in this formula.

However, the only case that I can think of that you can do that is, if you have that $$P(A)=1$$ and $$P(B)=0$$.

From the probability axioms, it is known that the probability of the entire space will be equal to $$1$$, i.e. $$P(\Omega)=1=P(A)$$.

We also know that $$P(\Omega)= 1- P(\Omega^{c})$$, where $$\Omega \cap \Omega^{c}=\varnothing$$. In this case $$P(\Omega^{c})=P(B)=0$$.

So, in this particular case we can identify that $$A\cap B=\varnothing \Rightarrow P(A\cap B)=0$$ they are independent.