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What is some intuitive insight regarding the conditional probability definition: $P(A\mid B) = \large \frac{P(A \cap B)}{P(B)}$ ? 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.

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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. – muzzlator Feb 4 '13 at 15:34
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 $A\cap B$ to the area of $B$. – David Mitra Feb 4 '13 at 15:35
up vote 11 down vote accepted

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 \[\mathbb P(A\mid B) = \frac{\text{# occurrences of A and B}}{\text{# occurrences of B}} = \frac{\mathbb P(A \cap B)}{\mathbb P(B)}\]

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 (as a commenter mentions, think Venn diagrams). So, for example, given that your roll result was even on a six-sided die, it's 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.

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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 $B$.

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