What is P(A|B) in Venn diagram
$begingroup$
I am trying to understand $P(A|B)$ in practice. I know that $(A,B)$ is the intersection between $A$ and $B$ on the diagram, and I know as well that $P(A|B)$ is probability of $A$ given $B$. But where is $P(A|B)$ on the diagram ?
probability conditional-probability
edited Jan 2 at 12:31
OmG
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
$endgroup$
add a comment |
$begingroup$
I am trying to understand $P(A|B)$ in practice. I know that $(A,B)$ is the intersection between $A$ and $B$ on the diagram, and I know as well that $P(A|B)$ is probability of $A$ given $B$. But where is $P(A|B)$ on the diagram ?
probability conditional-probability
edited Jan 2 at 12:31
OmG
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
$endgroup$
2
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34
add a comment |
$begingroup$
I am trying to understand $P(A|B)$ in practice. I know that $(A,B)$ is the intersection between $A$ and $B$ on the diagram, and I know as well that $P(A|B)$ is probability of $A$ given $B$. But where is $P(A|B)$ on the diagram ?
probability conditional-probability
edited Jan 2 at 12:31
OmG
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
$endgroup$
I am trying to understand $P(A|B)$ in practice. I know that $(A,B)$ is the intersection between $A$ and $B$ on the diagram, and I know as well that $P(A|B)$ is probability of $A$ given $B$. But where is $P(A|B)$ on the diagram ?
probability conditional-probability
probability conditional-probability
edited Jan 2 at 12:31
OmG
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
edited Jan 2 at 12:31
OmG
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
edited Jan 2 at 12:31
OmG
35629
edited Jan 2 at 12:31
OmG
35629
edited Jan 2 at 12:31
OmG
35629
35629
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
asked Jan 2 at 10:29
Mehdi SouregiMehdi Souregi
1115
1115
2
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34
add a comment |
2
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34
2
2
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the definition of $P(A|B)$ which is equal to $frac{P(Acap B)}{P(B)}$. Hence, you can't show it explicitly on the diagram as it is defined base on the division of two parts on the diagram. The value of blue part over the value of red circle. Hence it is $frac{0.1}{0.3 + 0.1} = frac{1}{4}$.
answered Jan 2 at 10:34
OmGOmG
35629
$endgroup$
add a comment |
$begingroup$
Remember that
$$
P(A|B) = frac{P(A, B)}{P(B)} = frac{0.1}{0.3 + 0.1} = frac{1}{4},
$$
which means that $P(A|B)$ is given by the proportion of the blue zone in your picture with respect to the red $B$ circle. This is not immediately visible in the diagram, so you'll have to use your imagination a bit to see the blue zone being $1/4$ of the size of the red circle.
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the definition of $P(A|B)$ which is equal to $frac{P(Acap B)}{P(B)}$. Hence, you can't show it explicitly on the diagram as it is defined base on the division of two parts on the diagram. The value of blue part over the value of red circle. Hence it is $frac{0.1}{0.3 + 0.1} = frac{1}{4}$.
answered Jan 2 at 10:34
OmGOmG
35629
$endgroup$
add a comment |
$begingroup$
Using the definition of $P(A|B)$ which is equal to $frac{P(Acap B)}{P(B)}$. Hence, you can't show it explicitly on the diagram as it is defined base on the division of two parts on the diagram. The value of blue part over the value of red circle. Hence it is $frac{0.1}{0.3 + 0.1} = frac{1}{4}$.
answered Jan 2 at 10:34
OmGOmG
35629
$endgroup$
add a comment |
$begingroup$
Using the definition of $P(A|B)$ which is equal to $frac{P(Acap B)}{P(B)}$. Hence, you can't show it explicitly on the diagram as it is defined base on the division of two parts on the diagram. The value of blue part over the value of red circle. Hence it is $frac{0.1}{0.3 + 0.1} = frac{1}{4}$.
answered Jan 2 at 10:34
OmGOmG
35629
$endgroup$
Using the definition of $P(A|B)$ which is equal to $frac{P(Acap B)}{P(B)}$. Hence, you can't show it explicitly on the diagram as it is defined base on the division of two parts on the diagram. The value of blue part over the value of red circle. Hence it is $frac{0.1}{0.3 + 0.1} = frac{1}{4}$.
answered Jan 2 at 10:34
OmGOmG
35629
answered Jan 2 at 10:34
OmGOmG
35629
answered Jan 2 at 10:34
OmGOmG
35629
answered Jan 2 at 10:34
OmGOmG
35629
35629
add a comment |
add a comment |
$begingroup$
Remember that
$$
P(A|B) = frac{P(A, B)}{P(B)} = frac{0.1}{0.3 + 0.1} = frac{1}{4},
$$
which means that $P(A|B)$ is given by the proportion of the blue zone in your picture with respect to the red $B$ circle. This is not immediately visible in the diagram, so you'll have to use your imagination a bit to see the blue zone being $1/4$ of the size of the red circle.
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
$endgroup$
add a comment |
$begingroup$
Remember that
$$
P(A|B) = frac{P(A, B)}{P(B)} = frac{0.1}{0.3 + 0.1} = frac{1}{4},
$$
which means that $P(A|B)$ is given by the proportion of the blue zone in your picture with respect to the red $B$ circle. This is not immediately visible in the diagram, so you'll have to use your imagination a bit to see the blue zone being $1/4$ of the size of the red circle.
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
$endgroup$
add a comment |
$begingroup$
Remember that
$$
P(A|B) = frac{P(A, B)}{P(B)} = frac{0.1}{0.3 + 0.1} = frac{1}{4},
$$
which means that $P(A|B)$ is given by the proportion of the blue zone in your picture with respect to the red $B$ circle. This is not immediately visible in the diagram, so you'll have to use your imagination a bit to see the blue zone being $1/4$ of the size of the red circle.
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
$endgroup$
Remember that
$$
P(A|B) = frac{P(A, B)}{P(B)} = frac{0.1}{0.3 + 0.1} = frac{1}{4},
$$
which means that $P(A|B)$ is given by the proportion of the blue zone in your picture with respect to the red $B$ circle. This is not immediately visible in the diagram, so you'll have to use your imagination a bit to see the blue zone being $1/4$ of the size of the red circle.
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
answered Jan 2 at 10:36
Waldir LeoncioWaldir Leoncio
1,45352235
1,45352235
add a comment |
add a comment |
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2
$begingroup$
It is not on the diagram.
$endgroup$
– Xi'an
Jan 2 at 10:30
$begingroup$
How can I represent it (on my mind) ?
$endgroup$
– Mehdi Souregi
Jan 2 at 10:32
$begingroup$
Is it a part of (A,B) or part of A, I am confused
$endgroup$
– Mehdi Souregi
Jan 2 at 10:34