# Conditional Probability

1. Nov 28, 2014

$P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1$
So given the the event "A and B" as the sample space, the probability of A occurring is 1.

$P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}$
Those two events are independent if and only if the probability of "A or B" occurring is 1, in which case the conditional probability of A equals the probability of A.

Is my reasoning correct?

2. Nov 28, 2014

### BruceW

you mean that A and A∪B are independent if P(A∪B)=1 ? yes, this is a special case of independent events. There is one other special case which can mean that A and A∪B are independent. You are trying to make the last equation go from
$$P(A|A∪B) = \frac{P(A)}{P(A∪B)}$$
to become:
$$P(A|A∪B) = P(A)$$
right? and you did this by setting P(A∪B)=1. But there is another way also.

3. Nov 29, 2014

### Stephen Tashi

We'd have to consider the case $B \cap A = \emptyset$

Which two events? $A$ and $A \cup B$ ?

Suppose $B = \emptyset$.

I think your results are interesting and worth perfecting by taking care of the exceptional cases.

4. Nov 30, 2014

What about $P(A|B)$? My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?