Mixing
Solving the problem of conditional probability without using P($A | B$).
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I was trying to differentiate $A|B$ and $A cap B$. I think understanding the following problem will clear my doubt.
There are two urns namely $U_1$ and $U_2$. $U_1$ contains three white balls and five red balls while $U_2$ contains four white balls and five red balls. One Urn is chosen randomly and a white ball is drawn. What is the probability of drawing the white ball from the $U_1$?
If I name the event of choosing the urn is $A$ and drawing a white ball is $B$ , Then what will be the event $A cap B$ and how the sample space will look like.
I want to determine P($A cap B$) by determining $ fracn(Acap B)n(S) $. $S$ denotes the sample space. I know how to solve it by using conditional probability. I can not see the sample space. Can anyone elaborately explain me what will be the sample space $S$. I want to solve this problem by the way Bram28 has solved here
Can anyone please help me to understand?
probability
asked Jul 30 at 15:27
cmi
5999
 |Â
show 3 more comments
up vote
2
down vote
favorite
I was trying to differentiate $A|B$ and $A cap B$. I think understanding the following problem will clear my doubt.
There are two urns namely $U_1$ and $U_2$. $U_1$ contains three white balls and five red balls while $U_2$ contains four white balls and five red balls. One Urn is chosen randomly and a white ball is drawn. What is the probability of drawing the white ball from the $U_1$?
If I name the event of choosing the urn is $A$ and drawing a white ball is $B$ , Then what will be the event $A cap B$ and how the sample space will look like.
I want to determine P($A cap B$) by determining $ fracn(Acap B)n(S) $. $S$ denotes the sample space. I know how to solve it by using conditional probability. I can not see the sample space. Can anyone elaborately explain me what will be the sample space $S$. I want to solve this problem by the way Bram28 has solved here
Can anyone please help me to understand?
probability
asked Jul 30 at 15:27
cmi
5999
1
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
1
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
2
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was trying to differentiate $A|B$ and $A cap B$. I think understanding the following problem will clear my doubt.
There are two urns namely $U_1$ and $U_2$. $U_1$ contains three white balls and five red balls while $U_2$ contains four white balls and five red balls. One Urn is chosen randomly and a white ball is drawn. What is the probability of drawing the white ball from the $U_1$?
If I name the event of choosing the urn is $A$ and drawing a white ball is $B$ , Then what will be the event $A cap B$ and how the sample space will look like.
I want to determine P($A cap B$) by determining $ fracn(Acap B)n(S) $. $S$ denotes the sample space. I know how to solve it by using conditional probability. I can not see the sample space. Can anyone elaborately explain me what will be the sample space $S$. I want to solve this problem by the way Bram28 has solved here
Can anyone please help me to understand?
probability
asked Jul 30 at 15:27
cmi
5999
I was trying to differentiate $A|B$ and $A cap B$. I think understanding the following problem will clear my doubt.
There are two urns namely $U_1$ and $U_2$. $U_1$ contains three white balls and five red balls while $U_2$ contains four white balls and five red balls. One Urn is chosen randomly and a white ball is drawn. What is the probability of drawing the white ball from the $U_1$?
If I name the event of choosing the urn is $A$ and drawing a white ball is $B$ , Then what will be the event $A cap B$ and how the sample space will look like.
I want to determine P($A cap B$) by determining $ fracn(Acap B)n(S) $. $S$ denotes the sample space. I know how to solve it by using conditional probability. I can not see the sample space. Can anyone elaborately explain me what will be the sample space $S$. I want to solve this problem by the way Bram28 has solved here
Can anyone please help me to understand?
probability
asked Jul 30 at 15:27
cmi
5999
edited Jul 30 at 16:29
asked Jul 30 at 15:27
cmi
5999
asked Jul 30 at 15:27
cmi
5999
asked Jul 30 at 15:27
cmi
5999
5999
1
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
1
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
2
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57
 |Â
show 3 more comments
1
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
1
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
2
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57
1
1
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
1
1
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
2
2
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57
 |Â
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3 Answers
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up vote
1
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You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(Acap B)$ (the probability of picking an urn and then picking a white ball) by finding $fracn(Acap B)n(S)$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(Acap B)neqfracn(Acap B)n(S).$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(Acap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So
$$fracn(Acap B)n(S)=frac1103.$$
But the probability of drawing the white ball from $U_1$ is $frac12timesfrac12=frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=fracP(Acap B)P(B).$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$Acap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=1,2,3,4,5,6$. Bram28 constructs two more events of interest: $A=4,5,6$ and $B=2,4,6$. He then calculates
$$
P(Acap B)
=fracn(Acap B)n(S)
=fracn(4,6)n(1,2,3,4,5,6)
=frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(Acap B)
=
P(4,6)
=
P(4)+P(6)
=
frac16+frac16
=
frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=(1,W),(1,R),(2,W_1),(2,W_2),dots,(2,W_100),(2,R).
$$
I said that Bram28's calculation method won't work for this example because
$$
fracn(Acap B)n(S)=fracn((1,W))103=frac1103.
$$
answered Jul 30 at 16:28
Babamots
1157
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
 |Â
show 1 more comment
up vote
0
down vote
Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = E_1, E_2, E_3, E_4$
where,
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball
$B^c$: Pick Black ball
$E_1 = Acap B, E_2 = A^c cap B, E_3 = A cap B^c, E_4 = A^c cap B^c$
($Acap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $fracn(Acap B)n(S)$ is not valid).
answered Jul 30 at 16:36
Suhan Shetty
835
add a comment |Â
up vote
-1
down vote
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answered Jul 30 at 17:47
user580606
1
yea I got you...……………….
– cmi
Jul 30 at 17:51
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(Acap B)$ (the probability of picking an urn and then picking a white ball) by finding $fracn(Acap B)n(S)$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(Acap B)neqfracn(Acap B)n(S).$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(Acap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So
$$fracn(Acap B)n(S)=frac1103.$$
But the probability of drawing the white ball from $U_1$ is $frac12timesfrac12=frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=fracP(Acap B)P(B).$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$Acap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=1,2,3,4,5,6$. Bram28 constructs two more events of interest: $A=4,5,6$ and $B=2,4,6$. He then calculates
$$
P(Acap B)
=fracn(Acap B)n(S)
=fracn(4,6)n(1,2,3,4,5,6)
=frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(Acap B)
=
P(4,6)
=
P(4)+P(6)
=
frac16+frac16
=
frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=(1,W),(1,R),(2,W_1),(2,W_2),dots,(2,W_100),(2,R).
$$
I said that Bram28's calculation method won't work for this example because
$$
fracn(Acap B)n(S)=fracn((1,W))103=frac1103.
$$
answered Jul 30 at 16:28
Babamots
1157
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
 |Â
show 1 more comment
up vote
1
down vote
You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(Acap B)$ (the probability of picking an urn and then picking a white ball) by finding $fracn(Acap B)n(S)$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(Acap B)neqfracn(Acap B)n(S).$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(Acap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So
$$fracn(Acap B)n(S)=frac1103.$$
But the probability of drawing the white ball from $U_1$ is $frac12timesfrac12=frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=fracP(Acap B)P(B).$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$Acap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=1,2,3,4,5,6$. Bram28 constructs two more events of interest: $A=4,5,6$ and $B=2,4,6$. He then calculates
$$
P(Acap B)
=fracn(Acap B)n(S)
=fracn(4,6)n(1,2,3,4,5,6)
=frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(Acap B)
=
P(4,6)
=
P(4)+P(6)
=
frac16+frac16
=
frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=(1,W),(1,R),(2,W_1),(2,W_2),dots,(2,W_100),(2,R).
$$
I said that Bram28's calculation method won't work for this example because
$$
fracn(Acap B)n(S)=fracn((1,W))103=frac1103.
$$
answered Jul 30 at 16:28
Babamots
1157
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(Acap B)$ (the probability of picking an urn and then picking a white ball) by finding $fracn(Acap B)n(S)$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(Acap B)neqfracn(Acap B)n(S).$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(Acap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So
$$fracn(Acap B)n(S)=frac1103.$$
But the probability of drawing the white ball from $U_1$ is $frac12timesfrac12=frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=fracP(Acap B)P(B).$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$Acap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=1,2,3,4,5,6$. Bram28 constructs two more events of interest: $A=4,5,6$ and $B=2,4,6$. He then calculates
$$
P(Acap B)
=fracn(Acap B)n(S)
=fracn(4,6)n(1,2,3,4,5,6)
=frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(Acap B)
=
P(4,6)
=
P(4)+P(6)
=
frac16+frac16
=
frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=(1,W),(1,R),(2,W_1),(2,W_2),dots,(2,W_100),(2,R).
$$
I said that Bram28's calculation method won't work for this example because
$$
fracn(Acap B)n(S)=fracn((1,W))103=frac1103.
$$
answered Jul 30 at 16:28
Babamots
1157
You call $A$ "the event of choosing an urn," but events need to be specific things. As you've stated it, $A=S$, meaning every event in your sample space involves picking an urn. That's just part of the problem setup.
Since you are looking for $P(A|B)$ and the problem is to find the "Probability of picking $U_1$ given that a white ball was drawn," you probably meant that $A$ is the event of picking $U_1$.
You say you want to find $P(Acap B)$ (the probability of picking an urn and then picking a white ball) by finding $fracn(Acap B)n(S)$ (the number of ways you can draw a white ball divided by the size of the sample space). You appear to believe that these quantities are equal, but as a general rule,
$$P(Acap B)neqfracn(Acap B)n(S).$$
You can only count on these being equal if each of the ways in which you could draw a white ball is equally likely to occur (for the urn problem, they're not). Let's change the numbers to make this obvious: if $U_1$ had one red, one white and $U_2$ had one red, 100 white, then $n(Acap B)=1$ because there is 1 way to draw the white ball form $U_1$. The total number of possible outcomes is the number of balls, $n(S)=103$. So
$$fracn(Acap B)n(S)=frac1103.$$
But the probability of drawing the white ball from $U_1$ is $frac12timesfrac12=frac14$
You say that you know how to do the problem with conditional probability so you don't want any conditional probability solutions. I am quite certain that there is no correct solution method that does not involve conditional probability. You can get rid of conditional probabilities quickly with Bayes' rule, of course:
$$P(A|B)=fracP(Acap B)P(B).$$
You might want to post your solution as a comment and ask for help checking its accuracy.
edit:
joriki is correct that $(A|B)$ is not an event. In fact, without talking about probability, $(A|B)$ doesn't really have a meaning. $P(A|B)$ is the probability of the event $A$, but measured in a different way. By putting "$|B$" in there, what you mean is that your experiment has fundamentally changed. $P(A|B)$ means you want the chances of $A$ (picking $U_1$) in an experiment where, if event $B$ does not occur (a red ball is drawn), you ignore the result and start over.
$Acap B$ is an event: the event that both event $A$ and event $B$ occur.
edit again:
You can't solve this problem in the same way as that example. In Bram28's example, there is a die being thrown, and there are 6 events that each have equal chances of occurring. These are $S=1,2,3,4,5,6$. Bram28 constructs two more events of interest: $A=4,5,6$ and $B=2,4,6$. He then calculates
$$
P(Acap B)
=fracn(Acap B)n(S)
=fracn(4,6)n(1,2,3,4,5,6)
=frac26
$$
This shortcut is only correct because each event has the same probability of occurring:
$$
P(Acap B)
=
P(4,6)
=
P(4)+P(6)
=
frac16+frac16
=
frac26.
$$
In my 103 ball example, the events in the sample space do not have equal chances. There is more than one way to write the sample space. I choose to write members of $S$ as pairs with the number of the chosen urn first and a letter for the color of the drawn ball second. I'm going to pretend that every white ball in $U_2$ has a number which will be a subscript, like this:
$$
S=(1,W),(1,R),(2,W_1),(2,W_2),dots,(2,W_100),(2,R).
$$
I said that Bram28's calculation method won't work for this example because
$$
fracn(Acap B)n(S)=fracn((1,W))103=frac1103.
$$
answered Jul 30 at 16:28
Babamots
1157
edited Jul 30 at 18:35
answered Jul 30 at 16:28
Babamots
1157
answered Jul 30 at 16:28
Babamots
1157
answered Jul 30 at 16:28
Babamots
1157
1157
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
 |Â
show 1 more comment
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
1
1
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
I did not get you. How $n(A cap B)$ is 101 ? @Babamots
– cmi
Jul 30 at 17:15
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
Sorry, I fixed that when I realized that you really must have meant that $A$ is the event of picking $U_1$. Refresh your page for a better answer.
– Babamots
Jul 30 at 17:18
1
1
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
My question still remains why 1/103 and 1/2 ? @Babamots
– cmi
Jul 30 at 17:57
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
I think you could not understand where my doubt is. Can you please check the link I have shared. I want to solve the problem in the same way Bram28 has done there.@Babamots
– cmi
Jul 30 at 18:01
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
@cmi I edited again to explain this more directly.
– Babamots
Jul 30 at 18:36
 |Â
show 1 more comment
up vote
0
down vote
Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = E_1, E_2, E_3, E_4$
where,
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball
$B^c$: Pick Black ball
$E_1 = Acap B, E_2 = A^c cap B, E_3 = A cap B^c, E_4 = A^c cap B^c$
($Acap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $fracn(Acap B)n(S)$ is not valid).
answered Jul 30 at 16:36
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = E_1, E_2, E_3, E_4$
where,
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball
$B^c$: Pick Black ball
$E_1 = Acap B, E_2 = A^c cap B, E_3 = A cap B^c, E_4 = A^c cap B^c$
($Acap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $fracn(Acap B)n(S)$ is not valid).
answered Jul 30 at 16:36
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = E_1, E_2, E_3, E_4$
where,
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball
$B^c$: Pick Black ball
$E_1 = Acap B, E_2 = A^c cap B, E_3 = A cap B^c, E_4 = A^c cap B^c$
($Acap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $fracn(Acap B)n(S)$ is not valid).
answered Jul 30 at 16:36
Suhan Shetty
835
Based on the events of interest you can consider the following sample space to describe your experiment:
$ S = E_1, E_2, E_3, E_4$
where,
$A$: Pick Urn-1
$A^c$ :Pick Urn-2
$B$ : Pick a white ball
$B^c$: Pick Black ball
$E_1 = Acap B, E_2 = A^c cap B, E_3 = A cap B^c, E_4 = A^c cap B^c$
($Acap B$ means "Pick a white ball from urn-1")
In particular, what you are trying to find is $P(E_1)$.
However, note that the elements of S have unequal weights (i.e, they are not equally likely). So, you can not use the naive definition of probability to find $P(E_1)$ (i.e., something like $fracn(Acap B)n(S)$ is not valid).
answered Jul 30 at 16:36
Suhan Shetty
835
edited Jul 30 at 18:23
answered Jul 30 at 16:36
Suhan Shetty
835
answered Jul 30 at 16:36
Suhan Shetty
835
answered Jul 30 at 16:36
Suhan Shetty
835
835
add a comment |Â
add a comment |Â
up vote
-1
down vote
We are all set I have E either ey and use the wishes everything is yes but even E everything eye text me eye roll eeryrhyfyr E we ye eye R Wrobleski I R really eyeyduduyr to E either way ye R to E run errands to run right by the kids are going to be able to get to my we ended up yet eye on right E eueudiey yes eyueeiy to be E either eye is E et you eyeyduduyr eye is ehufyiddh I just can't get over Wednesday unless E either ey eyueeiy I will say u we ye I ey ugh ye I email will E either way easier E it ends up being the kids ejdieydiue you can always be eye is better than nothing ye E everything is going out my E everything is
answered Jul 30 at 17:47
user580606
1
yea I got you...……………….
– cmi
Jul 30 at 17:51
add a comment |Â
up vote
-1
down vote
We are all set I have E either ey and use the wishes everything is yes but even E everything eye text me eye roll eeryrhyfyr E we ye eye R Wrobleski I R really eyeyduduyr to E either way ye R to E run errands to run right by the kids are going to be able to get to my we ended up yet eye on right E eueudiey yes eyueeiy to be E either eye is E et you eyeyduduyr eye is ehufyiddh I just can't get over Wednesday unless E either ey eyueeiy I will say u we ye I ey ugh ye I email will E either way easier E it ends up being the kids ejdieydiue you can always be eye is better than nothing ye E everything is going out my E everything is
answered Jul 30 at 17:47
user580606
1
yea I got you...……………….
– cmi
Jul 30 at 17:51
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
We are all set I have E either ey and use the wishes everything is yes but even E everything eye text me eye roll eeryrhyfyr E we ye eye R Wrobleski I R really eyeyduduyr to E either way ye R to E run errands to run right by the kids are going to be able to get to my we ended up yet eye on right E eueudiey yes eyueeiy to be E either eye is E et you eyeyduduyr eye is ehufyiddh I just can't get over Wednesday unless E either ey eyueeiy I will say u we ye I ey ugh ye I email will E either way easier E it ends up being the kids ejdieydiue you can always be eye is better than nothing ye E everything is going out my E everything is
answered Jul 30 at 17:47
user580606
1
We are all set I have E either ey and use the wishes everything is yes but even E everything eye text me eye roll eeryrhyfyr E we ye eye R Wrobleski I R really eyeyduduyr to E either way ye R to E run errands to run right by the kids are going to be able to get to my we ended up yet eye on right E eueudiey yes eyueeiy to be E either eye is E et you eyeyduduyr eye is ehufyiddh I just can't get over Wednesday unless E either ey eyueeiy I will say u we ye I ey ugh ye I email will E either way easier E it ends up being the kids ejdieydiue you can always be eye is better than nothing ye E everything is going out my E everything is
answered Jul 30 at 17:47
user580606
1
answered Jul 30 at 17:47
user580606
1
answered Jul 30 at 17:47
user580606
1
answered Jul 30 at 17:47
user580606
1
1
yea I got you...……………….
– cmi
Jul 30 at 17:51
add a comment |Â
yea I got you...……………….
– cmi
Jul 30 at 17:51
yea I got you...……………….
– cmi
Jul 30 at 17:51
yea I got you...……………….
– cmi
Jul 30 at 17:51
add a comment |Â
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1
Is there an event $Amid B$?
– Lord Shark the Unknown
Jul 30 at 15:33
I do not know. Then what do you call $(A|B)$?
– cmi
Jul 30 at 15:34
1
I don't: you used the notation $Amid B$ and you called it an event. Could you explain what you mean by it?
– Lord Shark the Unknown
Jul 30 at 15:35
If $B$ has already happened then what is the probability of happening $A$? answer will be P$(A|B)$.@LordSharktheUnknown
– cmi
Jul 30 at 15:43
2
@cmi: We know what $P(Amid B)$ means. (It has nothing to do with the temporal order of $A$ and $B$, by the way.) But you wrote about an "event" denoted by $Amid B$. No such event occurs in $P(Amid B)$.
– joriki
Jul 30 at 15:57