MHB How to Show Equality of Probabilities?

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The discussion revolves around demonstrating the equality of probabilities in a discrete probability space. Participants explore the relationship between events A and B, specifically focusing on the expression P(A∩B) - P(A)P(B) = P(A^c)P(B) - P(A^c∩B). They confirm that since A and A^c are disjoint, the total probability P(B) can be expressed as P(A∩B) + P(A^c∩B). The validity of specific probability values for B is also examined, concluding that P(B) cannot be 0.3 or 0.7 due to resulting negative probabilities or exceeding the total probability of 1. The conversation emphasizes the importance of understanding independence and the properties of probability measures.
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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:
 
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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 $P(A\cup A^c)=P(A)+P(A^c)$. 🤔
 
Klaas van Aarsen said:
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:
 
Yep. All correct. (Nod)
 
Klaas van Aarsen said:
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:
 
Yep. (Nod)p

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