Why $P(A+B)$ is equal to$ P(A)+P(B)−P(B|A)P(A)$ if $A$ and $B$ are dependent events?












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I understand why $P(A+B)=P(A)+P(B)−P(AB)$ for independent events. After all, if $A$ and $B$ share same outcomes (like event $A$ is that a die will show an even number, while $B$ is the same die showing a prime number. $2$ is a prime number, so they share outcome) then by just adding their probabilities we double count their shared outcomes (We need to remember that $A+B$ means probability that at the least one outcome of either $A$ or $B$ will take place. So shared outcomes cause double counting). We compensate it by using "$-P(AB)$", i.e. we take away probability that a shared outcome will happen.



Alas, but the same logic doesn't seem to work for dependent events, there is no need for shared outcomes for two events to be dependent. I just don't understand why we need "$−P(B|A)P(A)$" instead of "$-P(AB)$".



Even a hint would be appreciated.










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  • 3




    $begingroup$
    $P(Acap B)=P(Bmid A)P(A)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 5:10






  • 1




    $begingroup$
    It's true for all events.
    $endgroup$
    – Michael Rozenberg
    Jan 7 at 5:13










  • $begingroup$
    Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
    $endgroup$
    – user161005
    Jan 7 at 5:17
















0












$begingroup$


I understand why $P(A+B)=P(A)+P(B)−P(AB)$ for independent events. After all, if $A$ and $B$ share same outcomes (like event $A$ is that a die will show an even number, while $B$ is the same die showing a prime number. $2$ is a prime number, so they share outcome) then by just adding their probabilities we double count their shared outcomes (We need to remember that $A+B$ means probability that at the least one outcome of either $A$ or $B$ will take place. So shared outcomes cause double counting). We compensate it by using "$-P(AB)$", i.e. we take away probability that a shared outcome will happen.



Alas, but the same logic doesn't seem to work for dependent events, there is no need for shared outcomes for two events to be dependent. I just don't understand why we need "$−P(B|A)P(A)$" instead of "$-P(AB)$".



Even a hint would be appreciated.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $P(Acap B)=P(Bmid A)P(A)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 5:10






  • 1




    $begingroup$
    It's true for all events.
    $endgroup$
    – Michael Rozenberg
    Jan 7 at 5:13










  • $begingroup$
    Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
    $endgroup$
    – user161005
    Jan 7 at 5:17














0












0








0





$begingroup$


I understand why $P(A+B)=P(A)+P(B)−P(AB)$ for independent events. After all, if $A$ and $B$ share same outcomes (like event $A$ is that a die will show an even number, while $B$ is the same die showing a prime number. $2$ is a prime number, so they share outcome) then by just adding their probabilities we double count their shared outcomes (We need to remember that $A+B$ means probability that at the least one outcome of either $A$ or $B$ will take place. So shared outcomes cause double counting). We compensate it by using "$-P(AB)$", i.e. we take away probability that a shared outcome will happen.



Alas, but the same logic doesn't seem to work for dependent events, there is no need for shared outcomes for two events to be dependent. I just don't understand why we need "$−P(B|A)P(A)$" instead of "$-P(AB)$".



Even a hint would be appreciated.










share|cite|improve this question











$endgroup$




I understand why $P(A+B)=P(A)+P(B)−P(AB)$ for independent events. After all, if $A$ and $B$ share same outcomes (like event $A$ is that a die will show an even number, while $B$ is the same die showing a prime number. $2$ is a prime number, so they share outcome) then by just adding their probabilities we double count their shared outcomes (We need to remember that $A+B$ means probability that at the least one outcome of either $A$ or $B$ will take place. So shared outcomes cause double counting). We compensate it by using "$-P(AB)$", i.e. we take away probability that a shared outcome will happen.



Alas, but the same logic doesn't seem to work for dependent events, there is no need for shared outcomes for two events to be dependent. I just don't understand why we need "$−P(B|A)P(A)$" instead of "$-P(AB)$".



Even a hint would be appreciated.







probability






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edited Jan 7 at 11:10









Gnumbertester

19018




19018










asked Jan 7 at 5:09









user161005user161005

19112




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  • 3




    $begingroup$
    $P(Acap B)=P(Bmid A)P(A)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 5:10






  • 1




    $begingroup$
    It's true for all events.
    $endgroup$
    – Michael Rozenberg
    Jan 7 at 5:13










  • $begingroup$
    Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
    $endgroup$
    – user161005
    Jan 7 at 5:17














  • 3




    $begingroup$
    $P(Acap B)=P(Bmid A)P(A)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 5:10






  • 1




    $begingroup$
    It's true for all events.
    $endgroup$
    – Michael Rozenberg
    Jan 7 at 5:13










  • $begingroup$
    Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
    $endgroup$
    – user161005
    Jan 7 at 5:17








3




3




$begingroup$
$P(Acap B)=P(Bmid A)P(A)$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 5:10




$begingroup$
$P(Acap B)=P(Bmid A)P(A)$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 5:10




1




1




$begingroup$
It's true for all events.
$endgroup$
– Michael Rozenberg
Jan 7 at 5:13




$begingroup$
It's true for all events.
$endgroup$
– Michael Rozenberg
Jan 7 at 5:13












$begingroup$
Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
$endgroup$
– user161005
Jan 7 at 5:17




$begingroup$
Oops. Seems like I totally missed fact that P(AB)=P(B|A)P(A), silly me. So the formula is universal for all events, no matter is independent or not.
$endgroup$
– user161005
Jan 7 at 5:17










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        Sometimes a figure is worth a thousand words:



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        answered Jan 7 at 5:15









        David G. StorkDavid G. Stork

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