How to Show Equality of Probabilities?

[mathmari]

Hey! :giggle:

Let $(\Omega, p)$ be a discrete probability room with induced probability measure $P$ and let $A, B\subseteq \Omega$ be two events.
I want to show that $P(A\cap B)-P(A)P(B)=P(A^c)P(B)-P(A^c\cap B)$.

For that do we write to what for example $P(A^c\cap B)$ is equal to simplify the expression or which way is the best one? :unsure:

We have that $(A\cap B)\cap (A^c\cap B)=\emptyset$ and so \begin{align*}&P((A\cap B)\cup (A^c\cap B))=P(A\cap B)+P (A^c\cap B)\\ & \Rightarrow P((A\cup A^c)\cap B)=P(A\cap B)+P (A^c\cap B)\end{align*}
Now we have to show that $P((A\cup A^c)\cap B)=P(A)P(B)+P(A^c)P(B)=[P(A)+P(A^c)]P(B)$, right?
We get that result if $(A\cup A^c)$ and $B$ are independent, or not? How can we show that? :unsure:

Or is there an other (better) way to show the desired expression? :unsure:

 

[I like Serena]

Hey mathmari!

We have that $(A\cup A^c)=\Omega$ and $B\subseteq \Omega$.
So $(A\cup A^c)\cap B=B$.

Also note that a probability measure must have $P(\Omega)=1$.
And since $A$ and $A^c$ are disjoint, we also have $P(A\cup A^c)=P(A)+P(A^c)$.

 

[mathmari]

We have that $(A\cup A^c)=\Omega$ and $B\subseteq \Omega$.
So $(A\cup A^c)\cap B=B$.

Also note that a probability measure must have $P(\Omega)=1$.
And since $A$ and $A^c$ are disjoint, we also have $P(A\cup A^c)=P(A)+P(A^c)$.

So do we have the following ?

\begin{align*}&P((A\cap B)\cup (A^c\cap B))=P(A\cap B)+P (A^c\cap B) \\ & \Rightarrow P((A\cup A^c)\cap B)=P(A\cap B)+P (A^c\cap B)\\ &\Rightarrow P( B)=P(A\cap B)+P (A^c\cap B)\end{align*} and \begin{align*}P(A)P(B)+P(A^c)P(B)&=[P(A)+P(A^c)]P(B)\\ & =[P(A)+1-P(A))]P(B)\\ & =P(B)\end{align*} Combining these results we get \begin{align*}
&P(A\cap B)+P (A^c\cap B)=P(A)P(B)+P(A^c)P(B) \\ & \Rightarrow P(A\cap B)-P(A)P(B)=P(A^c)P(B)-P(A^c\cap B)\end{align*}
Is everything correct? :unsure:

 

[I like Serena]

Yep. All correct. (Nod)

 

[mathmari]

Yep. All correct. (Nod)

Great! (Sun)

Suppose we have that $P(A)=0.8$ and $P(A\cap B)=0.4$.

I want to check if $P(B)=0.3$ and $P(B)=0.7$ is possible.

For $P(B)=0.3$ : We substitute at the above proven equality and since we get then $P(A^c\cap B)=-0.1$ and since a probability cannot be negativ $P(B)$ cannot be $0.3$.

For $P(B)=0.7$ : Substituting at the above equality we get an acceptable probability. But substituting at $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ we get that $P(A\cup B)=1.1$ and since a probability cannot be greater than $1$ $P(B)$ cannot be $0.7$.

Is everything correct? :unsure:

 

[I like Serena]

Yep. (Nod)p

We can verify by drawing a Venn diagram. (Nerd)