Conditional probabilities summing to one
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Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.
Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$
probability conditional-probability
edited Dec 22 '18 at 11:21
Saad
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
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add a comment |
$begingroup$
Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.
Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$
probability conditional-probability
edited Dec 22 '18 at 11:21
Saad
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
$endgroup$
add a comment |
$begingroup$
Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.
Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$
probability conditional-probability
edited Dec 22 '18 at 11:21
Saad
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
$endgroup$
Let's assume that we have two Bernoulli random variables: $A$ (can be true or false) and $B$ (can be true or false), and further assume we have been given $P(A=text{true}mid B=text{true})$ and $P(A=text{true}mid B=text{false})$.
Is it possible to calculate $P(A=text{false}mid B=text{true})$ and $P(A=text{false}mid B=text{false})$ from this? I think what it must hold is that these four terms must sum to one, i.e. $$P(A=text{true}mid B=text{true}) + P(A=text{true}mid B=text{false}) \+ P(A=text{false}mid B=text{true}) + P(A=text{false}mid B=text{false}) = 1.$$
probability conditional-probability
probability conditional-probability
edited Dec 22 '18 at 11:21
Saad
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
edited Dec 22 '18 at 11:21
Saad
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
edited Dec 22 '18 at 11:21
Saad
20.6k92452
edited Dec 22 '18 at 11:21
Saad
20.6k92452
edited Dec 22 '18 at 11:21
Saad
20.6k92452
20.6k92452
asked Dec 22 '18 at 11:11
machinerymachinery
1365
asked Dec 22 '18 at 11:11
machinerymachinery
1365
asked Dec 22 '18 at 11:11
machinerymachinery
1365
1365
add a comment |
add a comment |
3 Answers
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$begingroup$
No it's incorrect.
$$P(A=true|B=true)+P(A=false|B=true)$$
$$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
$$=frac{P(B=true)}{P(B=true)}=1$$
Similarly,
$$P(A=true|B=false)+P(A=false|B=false)=1$$
Thus, yes you can calculate the variables required, but the equation you wrote was incorrect
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
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$begingroup$
Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$
but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
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$begingroup$
The summation you mention gives sum $2$ (not $1$).
This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$
In both cases if you know one of terms then you also know the other one.
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No it's incorrect.
$$P(A=true|B=true)+P(A=false|B=true)$$
$$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
$$=frac{P(B=true)}{P(B=true)}=1$$
Similarly,
$$P(A=true|B=false)+P(A=false|B=false)=1$$
Thus, yes you can calculate the variables required, but the equation you wrote was incorrect
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
No it's incorrect.
$$P(A=true|B=true)+P(A=false|B=true)$$
$$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
$$=frac{P(B=true)}{P(B=true)}=1$$
Similarly,
$$P(A=true|B=false)+P(A=false|B=false)=1$$
Thus, yes you can calculate the variables required, but the equation you wrote was incorrect
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
No it's incorrect.
$$P(A=true|B=true)+P(A=false|B=true)$$
$$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
$$=frac{P(B=true)}{P(B=true)}=1$$
Similarly,
$$P(A=true|B=false)+P(A=false|B=false)=1$$
Thus, yes you can calculate the variables required, but the equation you wrote was incorrect
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
$endgroup$
No it's incorrect.
$$P(A=true|B=true)+P(A=false|B=true)$$
$$=frac{P(A=true,B=true)}{P(B=true)}+frac{P(A=false,B=true)}{P(B=true)}$$
$$=frac{P(B=true)}{P(B=true)}=1$$
Similarly,
$$P(A=true|B=false)+P(A=false|B=false)=1$$
Thus, yes you can calculate the variables required, but the equation you wrote was incorrect
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 11:25
Ankit KumarAnkit Kumar
1,542221
1,542221
add a comment |
add a comment |
$begingroup$
Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$
but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$
but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$
but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
$endgroup$
Not quite. You have across all joint possibilities $$P(A=text{true} text{ and } B=text{true}) + P(A=text{true}text{ and }B=text{false}) \;,,+ P(A=text{false}text{ and } B=text{true}) + P(A=text{false}text{ and } B=text{false}) = 1$$
but for conditional probabilities you have two equations $$P(A=text{true}mid B=text{true}) + P(A=text{false}mid B=text{true}) =1\P(A=text{true}mid B=text{false}) + P(A=text{false}mid B=text{false}) = 1$$
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
answered Dec 22 '18 at 11:27
HenryHenry
101k482170
101k482170
add a comment |
add a comment |
$begingroup$
The summation you mention gives sum $2$ (not $1$).
This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$
In both cases if you know one of terms then you also know the other one.
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
The summation you mention gives sum $2$ (not $1$).
This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$
In both cases if you know one of terms then you also know the other one.
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
The summation you mention gives sum $2$ (not $1$).
This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$
In both cases if you know one of terms then you also know the other one.
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
$endgroup$
The summation you mention gives sum $2$ (not $1$).
This because: $$P(A=text{true}mid B=text{true})+P(A=text{false}mid B=text{true})=1$$ and: $$P(A=text{true}mid B=text{false})+P(A=text{false}mid B=text{false})=1$$
In both cases if you know one of terms then you also know the other one.
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
answered Dec 22 '18 at 11:33
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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