Mixing
Intuition behind the Definition of Conditional Probability (for 2 Events)
$begingroup$
What is some intuitive insight regarding the conditional probability definition: $P(Amid B) = large frac{P(A cap B)}{P(B)}$ ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask.
probability soft-question intuition
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
$endgroup$
add a comment |
$begingroup$
What is some intuitive insight regarding the conditional probability definition: $P(Amid B) = large frac{P(A cap B)}{P(B)}$ ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask.
probability soft-question intuition
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
$endgroup$
$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
2
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35
add a comment |
$begingroup$
What is some intuitive insight regarding the conditional probability definition: $P(Amid B) = large frac{P(A cap B)}{P(B)}$ ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask.
probability soft-question intuition
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
$endgroup$
What is some intuitive insight regarding the conditional probability definition: $P(Amid B) = large frac{P(A cap B)}{P(B)}$ ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask.
probability soft-question intuition
probability soft-question intuition
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
edited Feb 2 '16 at 19:03
Greek - Area 51 Proposal
3,176769105
3,176769105
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
asked Feb 4 '13 at 15:29
MackMack
2,0261766119
2,0261766119
$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
2
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35
add a comment |
$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
2
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35
$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
2
2
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).
Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.
Hence [mathbb P(Amid B) = frac{text{# occurrences of A and B}}{text{# occurrences of B}} = frac{mathbb P(A cap B)}{mathbb P(B)}]
because the "total number of possibilities" in the expressions for $mathbb P(B)$ and $mathbb P(A cap B)$ cancel.
Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers ${1,2,3,4,5,6}$ are less than $4$ but only a third of the numbers in our subsection ${2,4,6}$ are.
edited Jan 18 at 22:55
nbro
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
$endgroup$
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
add a comment |
$begingroup$
Let $Omega$ be your space of possibilities. Then $B$ is a subset of $Omega$. The probability $mathbb P$ induces a probability for events lying in $B$. Namely $mathbb P_B = frac{mathbb P}{mathbb P(B)}$. You can check the important fact that $mathbb P_B(B) = 1$.
Then the event $A cap B$ seen as an event in $B$ has probability $mathbb P_B(A cap B)$.
So I would say that this is just a base change from $Omega$ to $B$.
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
$endgroup$
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
$begingroup$
Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).
Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.
Hence [mathbb P(Amid B) = frac{text{# occurrences of A and B}}{text{# occurrences of B}} = frac{mathbb P(A cap B)}{mathbb P(B)}]
because the "total number of possibilities" in the expressions for $mathbb P(B)$ and $mathbb P(A cap B)$ cancel.
Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers ${1,2,3,4,5,6}$ are less than $4$ but only a third of the numbers in our subsection ${2,4,6}$ are.
edited Jan 18 at 22:55
nbro
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
$endgroup$
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
add a comment |
$begingroup$
Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).
Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.
Hence [mathbb P(Amid B) = frac{text{# occurrences of A and B}}{text{# occurrences of B}} = frac{mathbb P(A cap B)}{mathbb P(B)}]
because the "total number of possibilities" in the expressions for $mathbb P(B)$ and $mathbb P(A cap B)$ cancel.
Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers ${1,2,3,4,5,6}$ are less than $4$ but only a third of the numbers in our subsection ${2,4,6}$ are.
edited Jan 18 at 22:55
nbro
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
$endgroup$
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
add a comment |
$begingroup$
Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).
Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.
Hence [mathbb P(Amid B) = frac{text{# occurrences of A and B}}{text{# occurrences of B}} = frac{mathbb P(A cap B)}{mathbb P(B)}]
because the "total number of possibilities" in the expressions for $mathbb P(B)$ and $mathbb P(A cap B)$ cancel.
Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers ${1,2,3,4,5,6}$ are less than $4$ but only a third of the numbers in our subsection ${2,4,6}$ are.
edited Jan 18 at 22:55
nbro
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
$endgroup$
Consider probabilities as proportions. To say that something has probability one-sixth is to say it occurs one-sixth of the time (this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically). Often we calculate probabilities simply by dividing the number of possibilities in which our event of interest occurs, by the number of possibilities total – e.g. to calculate the odds of throwing an even number on a six-sided dice, we calculate $3/6$. (This works because each of the possibilities we are counting is equally likely, by assumption).
Now let's say we want to work out how often $A$ occurs, given that we know $B$ has occurred. Well, we need to find the occurrences of $A$ in this scenario, and divide by the total number of possibilities. When we know $B$ occurred, the occurrences of $A$ are all and exactly those situations in which both $A$ and $B$ occur, and since we're assuming $B$ occurred, the total number of possibilities are reduced to only those where that happened.
Hence [mathbb P(Amid B) = frac{text{# occurrences of A and B}}{text{# occurrences of B}} = frac{mathbb P(A cap B)}{mathbb P(B)}]
because the "total number of possibilities" in the expressions for $mathbb P(B)$ and $mathbb P(A cap B)$ cancel.
Essentially, what we are doing is focussing on a particular subsection of the potential events, and considering what proportion of that subsection satisfies whatever property you're interested in (think of Venn diagrams). So, for example, given that your roll result was even, on a six-sided die, it is less likely to be less than $4$, because half the numbers ${1,2,3,4,5,6}$ are less than $4$ but only a third of the numbers in our subsection ${2,4,6}$ are.
edited Jan 18 at 22:55
nbro
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
edited Jan 18 at 22:55
nbro
2,41663174
edited Jan 18 at 22:55
nbro
2,41663174
edited Jan 18 at 22:55
nbro
2,41663174
2,41663174
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
answered Feb 4 '13 at 16:13
Ben MillwoodBen Millwood
11.3k32049
11.3k32049
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
add a comment |
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
"this is only one interpretation: it suits our purposes and our intuition, so let's not worry too much about what it means philosophically" --- What does it mean philosophically?
$endgroup$
– WorldGov
Jul 21 '18 at 8:49
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
Well as I understand it the frequentist interpretation is to embed each event in some kind of series of many experiments and look at the long-run average occurrences of the outcomes. This is easy to do with coin flips or dice rolls but it's less clear what the appropriate series of experiments would be if you were doing something like asking what the probability was that the Cuban Missile Crisis would lead to nuclear war (say).
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:19
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
$begingroup$
The Bayesian interpretation sidesteps that problem by saying that probability is just a measure of your lack of information about what the outcome will be (and is therefore subject-dependent), and to say that something occurs with probability $p$ is just to say that you'd be willing (in some abstract idealized sense) to pay £$p$ (and no more) for the opportunity to win £1 if the thing happens.
$endgroup$
– Ben Millwood
Jul 22 '18 at 10:24
add a comment |
$begingroup$
Let $Omega$ be your space of possibilities. Then $B$ is a subset of $Omega$. The probability $mathbb P$ induces a probability for events lying in $B$. Namely $mathbb P_B = frac{mathbb P}{mathbb P(B)}$. You can check the important fact that $mathbb P_B(B) = 1$.
Then the event $A cap B$ seen as an event in $B$ has probability $mathbb P_B(A cap B)$.
So I would say that this is just a base change from $Omega$ to $B$.
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be your space of possibilities. Then $B$ is a subset of $Omega$. The probability $mathbb P$ induces a probability for events lying in $B$. Namely $mathbb P_B = frac{mathbb P}{mathbb P(B)}$. You can check the important fact that $mathbb P_B(B) = 1$.
Then the event $A cap B$ seen as an event in $B$ has probability $mathbb P_B(A cap B)$.
So I would say that this is just a base change from $Omega$ to $B$.
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be your space of possibilities. Then $B$ is a subset of $Omega$. The probability $mathbb P$ induces a probability for events lying in $B$. Namely $mathbb P_B = frac{mathbb P}{mathbb P(B)}$. You can check the important fact that $mathbb P_B(B) = 1$.
Then the event $A cap B$ seen as an event in $B$ has probability $mathbb P_B(A cap B)$.
So I would say that this is just a base change from $Omega$ to $B$.
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
$endgroup$
Let $Omega$ be your space of possibilities. Then $B$ is a subset of $Omega$. The probability $mathbb P$ induces a probability for events lying in $B$. Namely $mathbb P_B = frac{mathbb P}{mathbb P(B)}$. You can check the important fact that $mathbb P_B(B) = 1$.
Then the event $A cap B$ seen as an event in $B$ has probability $mathbb P_B(A cap B)$.
So I would say that this is just a base change from $Omega$ to $B$.
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
answered Feb 4 '13 at 15:36
Damien LDamien L
5,047933
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$begingroup$
Informally, the chances A occurs given that B has already occurred is taken by finding the probability that both A and B occur, and divide by the chance B has already occurred.
$endgroup$
– muzzlator
Feb 4 '13 at 15:34
2
$begingroup$
Drawing a Venn diagram for $A$ and $B$ and interpreting probabilities as areas may help. If you're given that $B$ has occurred, then you may restrict your attention to the new sample space $B$. The probability that $A$ occurs given $B$ is the ratio of the area of $Acap B$ to the area of $B$.
$endgroup$
– David Mitra
Feb 4 '13 at 15:35