Mutually exclusive vs. independent events: how to reconcile one vs. two experiments?
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This question has been already asked. Below for example a simplist proof.
Why independent events are never mutually exclusive?:
What puzzles me is that whether in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments. How this can be compared? For example, if two events are mutually exclusive, it is possible that first will come event A and then event B. So, they why they will not be independent?
probability
asked Nov 15 at 20:30
John
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up vote
0
down vote
favorite
This question has been already asked. Below for example a simplist proof.
Why independent events are never mutually exclusive?:
What puzzles me is that whether in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments. How this can be compared? For example, if two events are mutually exclusive, it is possible that first will come event A and then event B. So, they why they will not be independent?
probability
asked Nov 15 at 20:30
John
485
Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question has been already asked. Below for example a simplist proof.
Why independent events are never mutually exclusive?:
What puzzles me is that whether in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments. How this can be compared? For example, if two events are mutually exclusive, it is possible that first will come event A and then event B. So, they why they will not be independent?
probability
asked Nov 15 at 20:30
John
485
This question has been already asked. Below for example a simplist proof.
Why independent events are never mutually exclusive?:
What puzzles me is that whether in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments. How this can be compared? For example, if two events are mutually exclusive, it is possible that first will come event A and then event B. So, they why they will not be independent?
probability
probability
asked Nov 15 at 20:30
John
485
asked Nov 15 at 20:30
John
485
asked Nov 15 at 20:30
John
485
asked Nov 15 at 20:30
John
485
asked Nov 15 at 20:30
John
485
485
Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32
|
show 1 more comment
Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32
Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32
|
show 1 more comment
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Independent can make sense in the context of a single experiment. If you toss a penny and a dime, the events "the penny comes up $H$" and "the dime comes up $H$" are independent.
– lulu
Nov 15 at 20:32
"in case of mutually exclusive definition we deal with one experiment, in case of independent events we deal with two experiments" Incorrect. The rigorous definitions don't make any mention of experiments whatsoever. Two events $A,B$ in a probability space are said to be mutually exclusive iff $Acap B=emptyset$. Meanwhile two events $A,B$ in a probability space are said to be independent iff $Pr(Acap B)=Pr(A)Pr(B)$.
– JMoravitz
Nov 15 at 20:37
On the topic of how independent can make sense in a single experiment, if you object to lulu's example since there are two coins being flipped... consider instead the experiment where you roll one standard fair four-sided die. Let $A$ be the event that the number shown is even and let $B$ be the event that the number shown is prime. You have $Pr(A)=frac{|{2,4}|}{|{1,2,3,4}|} = frac{1}{2}$ and $Pr(B)=frac{|{2,3}|}{|{1,2,3,4}|}=frac{1}{2}$ while $Pr(Acap B) = frac{|{2}|}{|{1,2,3,4}|}=frac{1}{4}=Pr(A)Pr(B)$
– JMoravitz
Nov 15 at 20:47
On the flip side, you can have mutually exclusive events when there are multiple repetitions of an experiment, for example when flipping three coins you can have "The first coin is a head" is mutually exclusive from "The first coin is a tail" but is not mutually exclusive from "The second coin is a tail" (do not think that "head" is always mutually exclusive from "tail", that is not the case) Or you could have "The number of heads flipped is exactly 3" is mutually exclusive from "The number of heads flipped is exactly 1", etc...
– JMoravitz
Nov 15 at 20:49
Thanks for the answer! But what sense does it make in your first example with even and prime dice to calculate Pr(A)*Pr(B) when only one of them might happen? I do not see how can I interpret such thing...
– John
Nov 15 at 21:32