If two probability events are not disjoint, then are they independent for certain?

Question

Vice versa.

Why?

Thanks :)

For example, if P(A)=0.4, P(B)=0.5, and P(A or B)=0.7.

How would you interpret this?

Answer 1

This is not true.
If two events are not disjoint, then that means P(AnB) ≠ 0

Let's say P(A) = 0.5, P(B) = 0.5, P(AnB) = 0.4
This is possible and A and B are not disjoint.
But A and B are not independent because
P(AnB) ≠ P(A)*P(B)

If however A and B are disjoint, then they are certainly dependent, not independent. That's because if A occurs, then we know for sure that B has not occurred. So there is strong dependence.

*EDIT*
blue: P(A)=0.4, P(B)=0.5, and P(A or B)=0.7 is a very specific case.
In that particular case, A and B are independent. But that is not a general rule. It only works in very specific cases. Change that 0.7 to 0.69 and it does not work.

Answer 2

Uh, obviously not. Consider the following: let X be a uniform random variable on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then:

P(X≤5) = .5
P(X≥5) = .6
P((X≤5) ∩ (X≥5)) = P(X=5) = .1
P(X≤5) * P(X≥5) = .5*.6 = .3 ≠ .1

So although X≤5 and X≥5 are not disjoint, they are not independent. The converse is also false. Consider the same uniform random variable as above, then consider:

P(X=11) = 0
P(X=0) = 0
P((X=11)∩(X=0)) = 0
P(X=11) * P(X=0) = 0 = P((X=11)∩(X=0))

So in fact, these events are by definition independent, despite being disjoint. The example given above suggests a general theorem -- if A and B are two disjoint events, then A and B are independent iff one of them has probability zero.

Proof: Suppose A and B are disjoint. Then P(A∩B) = P(∅) = 0. By definition, A and B are independent iff P(A) * P(B) = P(A∩B) = 0, which happens iff either P(A) = 0 or P(B) = 0. Q.E.D.

Answer 3

if two events are not joined then they are for certain independent because they are not joined.