P[(B /C) U (C / B)]
what is the meaning of the expressions above?
is it equal P(B /C) + P (C / B)- P(C joint B)
and when P(C joint B)=0, P[(B /C) U (C / B)]=0
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P[(B /C) U (C / B)]
what is the meaning of the expressions above?
is it equal P(B /C) + P (C / B)- P(C joint B)
and when P(C joint B)=0, P[(B /C) U (C / B)]=0
Could it possibly beQuote:
Originally Posted by parklover
$\displaystyle \mathcal{P}[(B\setminus C)\cup (C\setminus B)]~?$
yes. you are right. aren't they the same thing?Quote:
Originally Posted by Plato
Often $\displaystyle B\setminus C$ is written $\displaystyle B-C$ and it means $\displaystyle B\cap \overline{C}$ where $\displaystyle \overline{C}$ is $\displaystyle C$ complement.Quote:
Originally Posted by parklover
So $\displaystyle \mathcal{P}[(B\setminus C)\cup (C\setminus B)]=\mathcal{P}(B)+\mathcal{P}(C)-2\mathcal{P}(B\cap C)$.
