For 3 independent events A, B and C, is $mathrm{P(A cap B cap C) = P(A)P(B)P(C)}$?
$begingroup$
For 3 independent events A, B and C
is $mathrm{P(A cap B cap C) = P(A)P(B)P(C)}$?
Just like for two independent events $mathrm{P(Acap B) = P(A)P(B)}$.
Intuitively I think it is because the denominator gives the total number of possible outcomes and the numerator gives the favourable outcomes.
Also, if we suppose that A,B and C are not independent and we keep changing the sample space after A is performed and then after B is performed, then would $mathrm{P(Acap Bcap C) = P(A)P(B)text{(with altered sample space)}P(C)text{(with twice altered sample space)}}$
I think I have used this in many problems and got the right answer but never thought about it rigorously.
Example problem:
A bag contains 1 red and 2 blue balls. An experiment consists of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour.
Probability that at least one blue ball is drawn is?
So to solve it we use $P(E) = 1- P(RRR) = 1- dfrac{1}{3}dfrac{2}{4}dfrac{3}{5} = 0.9$
Why did we multiply here? Didn't we use that $P(RRR)= P(R)P(R)P(R)$ ?
probability
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
$endgroup$
add a comment |
$begingroup$
For 3 independent events A, B and C
is $mathrm{P(A cap B cap C) = P(A)P(B)P(C)}$?
Just like for two independent events $mathrm{P(Acap B) = P(A)P(B)}$.
Intuitively I think it is because the denominator gives the total number of possible outcomes and the numerator gives the favourable outcomes.
Also, if we suppose that A,B and C are not independent and we keep changing the sample space after A is performed and then after B is performed, then would $mathrm{P(Acap Bcap C) = P(A)P(B)text{(with altered sample space)}P(C)text{(with twice altered sample space)}}$
I think I have used this in many problems and got the right answer but never thought about it rigorously.
Example problem:
A bag contains 1 red and 2 blue balls. An experiment consists of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour.
Probability that at least one blue ball is drawn is?
So to solve it we use $P(E) = 1- P(RRR) = 1- dfrac{1}{3}dfrac{2}{4}dfrac{3}{5} = 0.9$
Why did we multiply here? Didn't we use that $P(RRR)= P(R)P(R)P(R)$ ?
probability
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
$endgroup$
$begingroup$
You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
$endgroup$
– Akash Roy
Nov 24 '18 at 3:18
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It is correct upto two events only , you can easily verify this
$endgroup$
– Akash Roy
Nov 24 '18 at 3:19
$begingroup$
@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
$endgroup$
– Abcd
Nov 24 '18 at 4:10
$begingroup$
What you have written is only valid for a special case.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:30
$begingroup$
It is only valid when events are pairwise independent I guess.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:35
add a comment |
$begingroup$
For 3 independent events A, B and C
is $mathrm{P(A cap B cap C) = P(A)P(B)P(C)}$?
Just like for two independent events $mathrm{P(Acap B) = P(A)P(B)}$.
Intuitively I think it is because the denominator gives the total number of possible outcomes and the numerator gives the favourable outcomes.
Also, if we suppose that A,B and C are not independent and we keep changing the sample space after A is performed and then after B is performed, then would $mathrm{P(Acap Bcap C) = P(A)P(B)text{(with altered sample space)}P(C)text{(with twice altered sample space)}}$
I think I have used this in many problems and got the right answer but never thought about it rigorously.
Example problem:
A bag contains 1 red and 2 blue balls. An experiment consists of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour.
Probability that at least one blue ball is drawn is?
So to solve it we use $P(E) = 1- P(RRR) = 1- dfrac{1}{3}dfrac{2}{4}dfrac{3}{5} = 0.9$
Why did we multiply here? Didn't we use that $P(RRR)= P(R)P(R)P(R)$ ?
probability
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
$endgroup$
For 3 independent events A, B and C
is $mathrm{P(A cap B cap C) = P(A)P(B)P(C)}$?
Just like for two independent events $mathrm{P(Acap B) = P(A)P(B)}$.
Intuitively I think it is because the denominator gives the total number of possible outcomes and the numerator gives the favourable outcomes.
Also, if we suppose that A,B and C are not independent and we keep changing the sample space after A is performed and then after B is performed, then would $mathrm{P(Acap Bcap C) = P(A)P(B)text{(with altered sample space)}P(C)text{(with twice altered sample space)}}$
I think I have used this in many problems and got the right answer but never thought about it rigorously.
Example problem:
A bag contains 1 red and 2 blue balls. An experiment consists of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour.
Probability that at least one blue ball is drawn is?
So to solve it we use $P(E) = 1- P(RRR) = 1- dfrac{1}{3}dfrac{2}{4}dfrac{3}{5} = 0.9$
Why did we multiply here? Didn't we use that $P(RRR)= P(R)P(R)P(R)$ ?
probability
probability
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
edited Nov 24 '18 at 4:04
Abcd
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
asked Nov 24 '18 at 2:55
AbcdAbcd
3,02321135
3,02321135
$begingroup$
You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
$endgroup$
– Akash Roy
Nov 24 '18 at 3:18
$begingroup$
It is correct upto two events only , you can easily verify this
$endgroup$
– Akash Roy
Nov 24 '18 at 3:19
$begingroup$
@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
$endgroup$
– Abcd
Nov 24 '18 at 4:10
$begingroup$
What you have written is only valid for a special case.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:30
$begingroup$
It is only valid when events are pairwise independent I guess.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:35
add a comment |
$begingroup$
You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
$endgroup$
– Akash Roy
Nov 24 '18 at 3:18
$begingroup$
It is correct upto two events only , you can easily verify this
$endgroup$
– Akash Roy
Nov 24 '18 at 3:19
$begingroup$
@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
$endgroup$
– Abcd
Nov 24 '18 at 4:10
$begingroup$
What you have written is only valid for a special case.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:30
$begingroup$
It is only valid when events are pairwise independent I guess.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:35
$begingroup$
You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
$endgroup$
– Akash Roy
Nov 24 '18 at 3:18
$begingroup$
You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
$endgroup$
– Akash Roy
Nov 24 '18 at 3:18
$begingroup$
It is correct upto two events only , you can easily verify this
$endgroup$
– Akash Roy
Nov 24 '18 at 3:19
$begingroup$
It is correct upto two events only , you can easily verify this
$endgroup$
– Akash Roy
Nov 24 '18 at 3:19
$begingroup$
@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
$endgroup$
– Abcd
Nov 24 '18 at 4:10
$begingroup$
@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
$endgroup$
– Abcd
Nov 24 '18 at 4:10
$begingroup$
What you have written is only valid for a special case.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:30
$begingroup$
What you have written is only valid for a special case.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:30
$begingroup$
It is only valid when events are pairwise independent I guess.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:35
$begingroup$
It is only valid when events are pairwise independent I guess.
$endgroup$
– Akash Roy
Nov 24 '18 at 5:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
answered Nov 24 '18 at 4:04
NedNed
1,993910
$endgroup$
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
|
show 6 more comments
$begingroup$
If I remember right, the idea here is this:
$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$
Chain rule for conditional probability
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
$endgroup$
add a comment |
$begingroup$
If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:
A = getting heads on the first coin.
B = getting heads on the second coin.
C = getting the same outcome (heads or tails) on both coins.
Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(Acap B) = P(A)P(B) =1/4$, $P(Acap C) = P(A)P(C) = 1/4$, and $P(Bcap C) = P(B)P(C) = 1/4$, however $P(Acap Bcap C) = 1/4 neq P(A)P(B)P(C)$.
If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|Bcap C) = P(A)$, etc., then the claim is true:
$$
P(Acap Bcap C) = P(A|Bcap C) P(Bcap C) = P(A) P(B) P(C) ,.
$$
For the second part, I assume it depends on how the sample space is altered.
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
answered Nov 24 '18 at 4:04
NedNed
1,993910
$endgroup$
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
|
show 6 more comments
$begingroup$
The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
answered Nov 24 '18 at 4:04
NedNed
1,993910
$endgroup$
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
|
show 6 more comments
$begingroup$
The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
answered Nov 24 '18 at 4:04
NedNed
1,993910
$endgroup$
The definition of "{A,B,C} is independent" implies that the equation you wrote is true, as well the pairwise equations being true. As shown in one of the answers below, this is stronger just having the three pairwise independence equations hold, and it's also stronger than just having the equation you wrote hold, although I don't remember a counterexample. In general, independence for a finite set $S$ of events means, by definition, that the intersection/product equation holds for EVERY subset of $S$.
answered Nov 24 '18 at 4:04
NedNed
1,993910
edited Nov 24 '18 at 4:05
answered Nov 24 '18 at 4:04
NedNed
1,993910
answered Nov 24 '18 at 4:04
NedNed
1,993910
answered Nov 24 '18 at 4:04
NedNed
1,993910
1,993910
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
|
show 6 more comments
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
What does pairwise independence mean?
$endgroup$
– Abcd
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
It means "taken two at a time"
$endgroup$
– Ned
Nov 24 '18 at 4:05
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
Could you give a counterexample to when three events are independent but not pairwise independent?
$endgroup$
– Abcd
Nov 24 '18 at 4:06
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
$begingroup$
I'll think about it, I constructed one a few years ago when I was teaching probability but I don't remember it now -- but just to be accurate, in such an example the set S of three events is not independent, since that would require the product/intersection equality to hold for ALL subsets of S.
$endgroup$
– Ned
Nov 24 '18 at 4:11
1
1
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
$begingroup$
That example uses conditional probability, not independence, and the multiplication you are correctly using is what Akash Roy is calling the chain rule for probability. The event "2nd ball is R" is not independent of "1st ball red" but your equation works because the 2/4 in there is P(2nd Red | 1st Red), etc
$endgroup$
– Ned
Nov 24 '18 at 4:28
|
show 6 more comments
$begingroup$
If I remember right, the idea here is this:
$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$
Chain rule for conditional probability
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
$endgroup$
add a comment |
$begingroup$
If I remember right, the idea here is this:
$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$
Chain rule for conditional probability
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
$endgroup$
add a comment |
$begingroup$
If I remember right, the idea here is this:
$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$
Chain rule for conditional probability
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
$endgroup$
If I remember right, the idea here is this:
$P(A,B,C) = P(A|B,C)*P(B,C) = P(A|B,C)*P(B|C)*P(C)$
Chain rule for conditional probability
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
answered Nov 24 '18 at 3:09
MakinaMakina
1,158316
1,158316
add a comment |
add a comment |
$begingroup$
If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:
A = getting heads on the first coin.
B = getting heads on the second coin.
C = getting the same outcome (heads or tails) on both coins.
Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(Acap B) = P(A)P(B) =1/4$, $P(Acap C) = P(A)P(C) = 1/4$, and $P(Bcap C) = P(B)P(C) = 1/4$, however $P(Acap Bcap C) = 1/4 neq P(A)P(B)P(C)$.
If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|Bcap C) = P(A)$, etc., then the claim is true:
$$
P(Acap Bcap C) = P(A|Bcap C) P(Bcap C) = P(A) P(B) P(C) ,.
$$
For the second part, I assume it depends on how the sample space is altered.
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
$begingroup$
If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:
A = getting heads on the first coin.
B = getting heads on the second coin.
C = getting the same outcome (heads or tails) on both coins.
Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(Acap B) = P(A)P(B) =1/4$, $P(Acap C) = P(A)P(C) = 1/4$, and $P(Bcap C) = P(B)P(C) = 1/4$, however $P(Acap Bcap C) = 1/4 neq P(A)P(B)P(C)$.
If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|Bcap C) = P(A)$, etc., then the claim is true:
$$
P(Acap Bcap C) = P(A|Bcap C) P(Bcap C) = P(A) P(B) P(C) ,.
$$
For the second part, I assume it depends on how the sample space is altered.
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
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add a comment |
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If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:
A = getting heads on the first coin.
B = getting heads on the second coin.
C = getting the same outcome (heads or tails) on both coins.
Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(Acap B) = P(A)P(B) =1/4$, $P(Acap C) = P(A)P(C) = 1/4$, and $P(Bcap C) = P(B)P(C) = 1/4$, however $P(Acap Bcap C) = 1/4 neq P(A)P(B)P(C)$.
If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|Bcap C) = P(A)$, etc., then the claim is true:
$$
P(Acap Bcap C) = P(A|Bcap C) P(Bcap C) = P(A) P(B) P(C) ,.
$$
For the second part, I assume it depends on how the sample space is altered.
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
$endgroup$
If the 3 events are just pairwise independent then it is easy to find counterexamples to the product rule. For instance consider the experiment "toss two coins", and the following events:
A = getting heads on the first coin.
B = getting heads on the second coin.
C = getting the same outcome (heads or tails) on both coins.
Those events are pairwise independent, we have $P(A) = P(B) = P(C) = 1/2$, and we have $P(Acap B) = P(A)P(B) =1/4$, $P(Acap C) = P(A)P(C) = 1/4$, and $P(Bcap C) = P(B)P(C) = 1/4$, however $P(Acap Bcap C) = 1/4 neq P(A)P(B)P(C)$.
If what you mean is that besides being pairwise independent each event is also independent from the other two, i.e., $P(A|Bcap C) = P(A)$, etc., then the claim is true:
$$
P(Acap Bcap C) = P(A|Bcap C) P(Bcap C) = P(A) P(B) P(C) ,.
$$
For the second part, I assume it depends on how the sample space is altered.
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
answered Nov 24 '18 at 3:49
mlerma54mlerma54
1,162138
1,162138
add a comment |
add a comment |
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You need to apply the chain rule of conditional probability here , the identity which you have stated is wrong.
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– Akash Roy
Nov 24 '18 at 3:18
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It is correct upto two events only , you can easily verify this
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– Akash Roy
Nov 24 '18 at 3:19
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@AkashRoy Are you sure I am wrong: math.stackexchange.com/a/2891712/476145 ?
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– Abcd
Nov 24 '18 at 4:10
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What you have written is only valid for a special case.
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– Akash Roy
Nov 24 '18 at 5:30
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It is only valid when events are pairwise independent I guess.
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– Akash Roy
Nov 24 '18 at 5:35