Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is A complement. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A and complement ASolution Check: Union/intersect complementThe Union of $n$ Independent Events Equals the Complement of the Complement of Their ProductProve that if $X$ and $Y$ are sets where $,left|Xright|=left|Yright|,,$ then $,left|Pleft(Xright)right|=left| Pleft(Yright)right|$.$Pleft((Acup B)(Acup B^complement)(A^complementcup B)right)$ has the valueComplement and IndependenceLet $O_1, O_2$ be independent events. Does the sigma algebras generated by $O_1$ and $O_1,O_2$ are independentFind $Pleft( 2M > X_1 + cdots + X_nright)$If $A$ and $B$ are any two events is it true that $Pleft(A|Bright)+Pleft(A|B^complementright)=1$Bijection $F= f in A to B to left(A/E to Bright)$
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Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is A complement.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A and complement ASolution Check: Union/intersect complementThe Union of $n$ Independent Events Equals the Complement of the Complement of Their ProductProve that if $X$ and $Y$ are sets where $,left|Xright|=left|Yright|,,$ then $,left|Pleft(Xright)right|=left| Pleft(Yright)right|$.$Pleft((Acup B)(Acup B^complement)(A^complementcup B)right)$ has the valueComplement and IndependenceLet $O_1, O_2$ be independent events. Does the sigma algebras generated by $O_1$ and $O_1,O_2$ are independentFind $Pleft( 2M > X_1 + cdots + X_nright)$If $A$ and $B$ are any two events is it true that $Pleft(A|Bright)+Pleft(A|B^complementright)=1$Bijection $F= forall left(<x, y> in Eright). to left(fleft(xright)=fleft(yright)right) to left(A/E to Bright)$
$begingroup$
Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is complement of set $A$. Are $A$ and $B $ independent sets?
I think not but I can not find example that shows so.
probability discrete-mathematics
asked Mar 31 at 14:25
user560461user560461
1618
$endgroup$
add a comment |
$begingroup$
Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is complement of set $A$. Are $A$ and $B $ independent sets?
I think not but I can not find example that shows so.
probability discrete-mathematics
asked Mar 31 at 14:25
user560461user560461
1618
$endgroup$
add a comment |
$begingroup$
Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is complement of set $A$. Are $A$ and $B $ independent sets?
I think not but I can not find example that shows so.
probability discrete-mathematics
asked Mar 31 at 14:25
user560461user560461
1618
$endgroup$
Let $P left( A right) > 0$ and $ P left( B | A right) = P left( B | C right) $ where $ C $ is complement of set $A$. Are $A$ and $B $ independent sets?
I think not but I can not find example that shows so.
probability discrete-mathematics
probability discrete-mathematics
asked Mar 31 at 14:25
user560461user560461
1618
asked Mar 31 at 14:25
user560461user560461
1618
edited Mar 31 at 14:35
user560461
asked Mar 31 at 14:25
user560461user560461
1618
asked Mar 31 at 14:25
user560461user560461
1618
asked Mar 31 at 14:25
user560461user560461
1618
1618
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Yes the are! We have $$P(Bcap A)over P(A)= P(Bcap A')over P(A')$$
Since $P(A')= 1-P(A)$ and $$P(Bcap A)+P(Bcap A')= P(B)$$ we have
$$P(Bcap A)+P(Bcap A)1-P(A)over P(A) = P(B)$$ From here we get $$P(Bcap A) = P(B)cdot P(A)$$
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes the are! We have $$P(Bcap A)over P(A)= P(Bcap A')over P(A')$$
Since $P(A')= 1-P(A)$ and $$P(Bcap A)+P(Bcap A')= P(B)$$ we have
$$P(Bcap A)+P(Bcap A)1-P(A)over P(A) = P(B)$$ From here we get $$P(Bcap A) = P(B)cdot P(A)$$
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
add a comment |
$begingroup$
Yes the are! We have $$P(Bcap A)over P(A)= P(Bcap A')over P(A')$$
Since $P(A')= 1-P(A)$ and $$P(Bcap A)+P(Bcap A')= P(B)$$ we have
$$P(Bcap A)+P(Bcap A)1-P(A)over P(A) = P(B)$$ From here we get $$P(Bcap A) = P(B)cdot P(A)$$
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
add a comment |
$begingroup$
Yes the are! We have $$P(Bcap A)over P(A)= P(Bcap A')over P(A')$$
Since $P(A')= 1-P(A)$ and $$P(Bcap A)+P(Bcap A')= P(B)$$ we have
$$P(Bcap A)+P(Bcap A)1-P(A)over P(A) = P(B)$$ From here we get $$P(Bcap A) = P(B)cdot P(A)$$
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
Yes the are! We have $$P(Bcap A)over P(A)= P(Bcap A')over P(A')$$
Since $P(A')= 1-P(A)$ and $$P(Bcap A)+P(Bcap A')= P(B)$$ we have
$$P(Bcap A)+P(Bcap A)1-P(A)over P(A) = P(B)$$ From here we get $$P(Bcap A) = P(B)cdot P(A)$$
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
answered Mar 31 at 14:31
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
add a comment |
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
That is exactly how I solved this, but my professor said opposite, so I wanted to check.
$endgroup$
– user560461
Mar 31 at 14:33
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
It could go wrong only if $P(A)=1$
$endgroup$
– Maria Mazur
Mar 31 at 14:34
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
$begingroup$
agreed. Thanks :)
$endgroup$
– user560461
Mar 31 at 14:35
add a comment |
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