# Conditional Probability

##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?

BruceW
Homework Helper
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.

Stephen Tashi
##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.
We'd have to consider the case $B \cap A = \emptyset$

##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
Those two events are independent
Which two events? $A$ and $A \cup B$ ?

if and only if the probability of "A or B" occurring is 1

Suppose $B = \emptyset$.

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

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.

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?

Stephen Tashi