Possibilities / probabilities
- Thread starter Barbra
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
To Barbra. I realize that you are working from Arabic text. But this is an English language website.
We use standard Western notation for mathematics here. Therefore because all of probability built
upon set-theory that is what we use here.
So in a probability space [imath]\mathit{S}\text{ the events are elements of the power set, }\mathcal{P}(\mathit{S})[/imath].
If [imath]\bf{A,~B,~C, \& X}[/imath] are each events in [imath]\mathit{S}[/imath] as such the probabilities are [imath]0\le \mathit{P}(\bf X )\le 1. [/imath]
Moreover, if [imath]\bf{A\subseteq B}\text{ then }\mathit{P}(\bf {A})\le\mathit{P}(\bf{B})[/imath]
We know that (**)[imath]{\bf A}\subseteq {\bf A\cup B}\;\;\therefore\;\mathit{P}(\bf {A})\le\mathit{P}(\bf {A\cup B})[/imath]
As I read the attachment, it clearly says
[imath]\mathit{P}(\bf A )=0.2 [/imath]
[imath]\mathit{P}(\bf B )=0.6[/imath]
[imath]\mathit{P}(\bf B^c\cap A^c )=0.7[/imath]
[imath]\mathit{P}(\bf A\cup B )=0.3[/imath]
Do you see a contradiction to (**) in your posting?
On another notational matter. there is no such [imath]\bf X/Y[/imath]
There is [imath]\bf X\setminus Y[/imath] meaning [imath]\bf X\cap Y^c[/imath]
[imath][/imath][imath][/imath][imath][/imath]
We use standard Western notation for mathematics here. Therefore because all of probability built
upon set-theory that is what we use here.
So in a probability space [imath]\mathit{S}\text{ the events are elements of the power set, }\mathcal{P}(\mathit{S})[/imath].
If [imath]\bf{A,~B,~C, \& X}[/imath] are each events in [imath]\mathit{S}[/imath] as such the probabilities are [imath]0\le \mathit{P}(\bf X )\le 1. [/imath]
Moreover, if [imath]\bf{A\subseteq B}\text{ then }\mathit{P}(\bf {A})\le\mathit{P}(\bf{B})[/imath]
We know that (**)[imath]{\bf A}\subseteq {\bf A\cup B}\;\;\therefore\;\mathit{P}(\bf {A})\le\mathit{P}(\bf {A\cup B})[/imath]
As I read the attachment, it clearly says
[imath]\mathit{P}(\bf A )=0.2 [/imath]
[imath]\mathit{P}(\bf B )=0.6[/imath]
[imath]\mathit{P}(\bf B^c\cap A^c )=0.7[/imath]
[imath]\mathit{P}(\bf A\cup B )=0.3[/imath]
Do you see a contradiction to (**) in your posting?
On another notational matter. there is no such [imath]\bf X/Y[/imath]
There is [imath]\bf X\setminus Y[/imath] meaning [imath]\bf X\cap Y^c[/imath]
[imath][/imath][imath][/imath][imath][/imath]
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,089
I've been assuming that P(A/B) is a misreading of P(A|B); in fact, this is one reason I asked to see the original, in order to check whether the notation used there might use the slash (though it seems unlikely).
