Mixing
Why $P(A+B)$ is equal to$ P(A)+P(B)−P(B|A)P(A)$ if $A$ and $B$ are dependent events?
$begingroup$
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.
probability
edited Jan 7 at 11:10
Gnumbertester
435111
asked Jan 7 at 5:09
user161005user161005
19112
$endgroup$
add a comment |
$begingroup$
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.
probability
edited Jan 7 at 11:10
Gnumbertester
435111
asked Jan 7 at 5:09
user161005user161005
19112
$endgroup$
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
add a comment |
$begingroup$
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.
probability
edited Jan 7 at 11:10
Gnumbertester
435111
asked Jan 7 at 5:09
user161005user161005
19112
$endgroup$
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
probability
edited Jan 7 at 11:10
Gnumbertester
435111
asked Jan 7 at 5:09
user161005user161005
19112
edited Jan 7 at 11:10
Gnumbertester
435111
asked Jan 7 at 5:09
user161005user161005
19112
edited Jan 7 at 11:10
Gnumbertester
435111
edited Jan 7 at 11:10
Gnumbertester
435111
edited Jan 7 at 11:10
Gnumbertester
435111
435111
asked Jan 7 at 5:09
user161005user161005
19112
asked Jan 7 at 5:09
user161005user161005
19112
asked Jan 7 at 5:09
user161005user161005
19112
19112
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Sometimes a figure is worth a thousand words:
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
add a comment |
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$begingroup$
Sometimes a figure is worth a thousand words:
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
Sometimes a figure is worth a thousand words:
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
answered Jan 7 at 5:15
David G. StorkDavid G. Stork
10.8k31432
10.8k31432
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add a comment |
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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