Joint probabilities of $A$ and independent $B$ and $C$
1
$begingroup$
Given that $B$ and $C$ are independent, when would $$P(Acap Bcap C) = P(Acap B)P(Acap C);?$$
probability
edited Dec 31 '18 at 21:18
greedoid
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
$endgroup$
add a comment |
1
$begingroup$
Given that $B$ and $C$ are independent, when would $$P(Acap Bcap C) = P(Acap B)P(Acap C);?$$
probability
edited Dec 31 '18 at 21:18
greedoid
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
$endgroup$
1
$begingroup$
A, B, C are threeeventsorlogical statements. comma representslogical andorconjunction. P(X) representsprobability that statement X is true.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
1
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53
add a comment |
1
1
1
$begingroup$
Given that $B$ and $C$ are independent, when would $$P(Acap Bcap C) = P(Acap B)P(Acap C);?$$
probability
edited Dec 31 '18 at 21:18
greedoid
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
$endgroup$
Given that $B$ and $C$ are independent, when would $$P(Acap Bcap C) = P(Acap B)P(Acap C);?$$
probability
probability
edited Dec 31 '18 at 21:18
greedoid
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
edited Dec 31 '18 at 21:18
greedoid
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
edited Dec 31 '18 at 21:18
greedoid
39k114797
edited Dec 31 '18 at 21:18
greedoid
39k114797
edited Dec 31 '18 at 21:18
greedoid
39k114797
39k114797
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
asked Dec 31 '18 at 19:40
hamster on wheelshamster on wheels
1155
1155
1
$begingroup$
A, B, C are threeeventsorlogical statements. comma representslogical andorconjunction. P(X) representsprobability that statement X is true.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
1
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53
add a comment |
1
$begingroup$
A, B, C are threeeventsorlogical statements. comma representslogical andorconjunction. P(X) representsprobability that statement X is true.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
1
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53
1
1
$begingroup$
A, B, C are three
events or logical statements. comma represents logical and or conjunction. P(X) represents probability that statement X is true.$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
A, B, C are three
events or logical statements. comma represents logical and or conjunction. P(X) represents probability that statement X is true.$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
1
1
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53
add a comment |
1 Answer
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oldest
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1
$begingroup$
We have
$$P(Acap Bcap C) = P(Bcap C)P(A|Bcap C) = P(B)P(C)P(A|Bcap C) $$
Since $$ P(Acap B)P(Acap C) = P(B)P(A|B)P(C)P(A|C)$$
we get $$boxed{P(A|Bcap C) = P(A|B)P(A|C)}$$
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
$endgroup$
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
We have
$$P(Acap Bcap C) = P(Bcap C)P(A|Bcap C) = P(B)P(C)P(A|Bcap C) $$
Since $$ P(Acap B)P(Acap C) = P(B)P(A|B)P(C)P(A|C)$$
we get $$boxed{P(A|Bcap C) = P(A|B)P(A|C)}$$
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
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1
$begingroup$
We have
$$P(Acap Bcap C) = P(Bcap C)P(A|Bcap C) = P(B)P(C)P(A|Bcap C) $$
Since $$ P(Acap B)P(Acap C) = P(B)P(A|B)P(C)P(A|C)$$
we get $$boxed{P(A|Bcap C) = P(A|B)P(A|C)}$$
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
$endgroup$
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1
1
1
$begingroup$
We have
$$P(Acap Bcap C) = P(Bcap C)P(A|Bcap C) = P(B)P(C)P(A|Bcap C) $$
Since $$ P(Acap B)P(Acap C) = P(B)P(A|B)P(C)P(A|C)$$
we get $$boxed{P(A|Bcap C) = P(A|B)P(A|C)}$$
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
$endgroup$
We have
$$P(Acap Bcap C) = P(Bcap C)P(A|Bcap C) = P(B)P(C)P(A|Bcap C) $$
Since $$ P(Acap B)P(Acap C) = P(B)P(A|B)P(C)P(A|C)$$
we get $$boxed{P(A|Bcap C) = P(A|B)P(A|C)}$$
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
edited Dec 31 '18 at 21:02
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
answered Dec 31 '18 at 20:56
greedoidgreedoid
39k114797
39k114797
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add a comment |
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1
$begingroup$
A, B, C are three
eventsorlogical statements. comma representslogical andorconjunction. P(X) representsprobability that statement X is true.$endgroup$
– hamster on wheels
Dec 31 '18 at 19:42
$begingroup$
Probably it requires that B and C never happen at the same time. If that is the case, the intersection of A, B, and C is same as the union of the (intersection of A and B) and the (intersection of B and C). wait... my brain is not working today.
$endgroup$
– hamster on wheels
Dec 31 '18 at 19:51
1
$begingroup$
That would require either $B$ or $C$ to be impossible.
$endgroup$
– copper.hat
Dec 31 '18 at 19:53