Hi
I was wondering, how do check whether the following is independent or not $\displaystyle (A \cup B) \cap (A' \cap B')$?
Thanks
As stated, this is a meaningless question.
It is true that $(A \cup B)~ \&~ (A^\prime \cap B^\prime)$ are mutually exclusive.
Assuming that the $^\prime$ is a notation for complement, the above is misstaken.
$ \begin{align*}(A \cup B) \cap (A^\prime \cap B^\prime)&=(A \cup B) \cap (A\cup B)^\prime \\&=\emptyset \end{align*}$
Thank you for your replies guys
Basically, I wanted to investigate whether $\displaystyle (A' \cap B') \& (A \cup B)$ are independent and or mutually exclusive.
I understand that for them to be mutually exclusive, the following condition must be satifised: $\displaystyle (A' \cap B') \cap (A \cup B) = \emptyset$
however for them to be independent, it must satisfy $\displaystyle P(A' \cap B') \cup P(A \cup B) =P(A' \cap B')*P(A \cup B)$
You guys mention, it's mutually exclusive, but how can I justify that.
That is what I posted. Two sets are mutually exclusive if their intersection is null.
You seem to be asking or thinking of events in a probability space? However, you gave no information whatever about the probability sets and/or probability measure.
The point is that ordinary sets can be said to be mutually exclusive but the concept of independence is not applied in general. In vector space certain vectors can be independent of one another.
Now suppose we have a probability space. The subsets of the space are the events.
Events $A~\&~B$ are mutually exclusive provided: $A\cap B=\emptyset,~\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)~\&~\mathcal{P}(A \cap B)=0$
Events $A~\&~B$ are independent provided: $\mathcal{P}(A\cap B)=\mathcal{P}(A)\cdot\mathcal{P}( B)$