MHB How to Show Equality of Probabilities?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Probabilities
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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:
 
Last edited by a moderator:
Physics news on Phys.org
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)
 
Last edited:
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.

Similar threads

Back
Top