# Why is the conditional probability formula not the intersection of A and B over A, rather than the intersection over P(A)?

The conditional probability of $$A$$ given the occurrence of $$B$$ is how likely $$A$$ is to have occurred given that $$B$$ has occurred. The formal definition of the conditional probability of $$A$$ given $$B$$ is $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

I get the numerator term: when we reduce the sample space to just occurrences of $$B$$, then the occurrences of $$A$$ will be all and only those occurrences of $$A$$ which are also occurrences of $$B$$ - i.e. the intersection of $$A$$ and $$B$$.

I don't get the denominator term, though. If we know that $$A$$ has occurred, then $$B$$ becomes the sample space. The probability of an event is defined to be the long-run relative frequency of favourable outcomes to all possible outcomes (i.e. the sample space). That is: $$P(A) = \frac{A}{\Omega}$$

But if $$B$$ becomes the sample space, then why are we assigning a probability to $$B$$, as opposed to dividing by the number of occurrences of $$B$$? In other words, should we not have: $$P(A|B)=\frac{A\cap B}{B}$$

The probability of $$B$$ here is just $$1$$, is it not? What have I missed?

• No, $B$ is a subset of the sample space. We are figuring out how much of $A$ is in $B$. Jun 23, 2022 at 15:29
• This is hard to follow. Counting measure is one type of probability distribution, but it is not the only type. In general, it is not clear what an expression like $\frac {A\cap B}{B}$ might mean.
– lulu
Jun 23, 2022 at 15:33
• There is no need to be rude. Jun 23, 2022 at 15:37

In the approach you are thinking of where you are "counting the number of favourable outcomes", it should really be the cardinality or size of the set: i.e. you $$\mathbb{P}(A) = \frac{|A|}{|\Omega|}$$ (where these all make sense). Now you suggest the conditional probability should be of the form $$\mathbb{P}(A \mid B) = \frac{|A \cap B|}{|B|}$$; compare this to the more general definition $$\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$. These are essentially the same thing, when our probabilities are defined by counting, i.e. $$\mathbb{P}(A \cap B) = \frac{|A \cap B|}{|\Omega|}$$ and $$\mathbb{P}(B) = \frac{|B|}{|\Omega|}$$ (sub them in!)

Recall that $$P(A|B)P(B) = P(B|A)P(A) = P(A \cap B)$$.

In other words, the probability of A is conditional upon B occurring, meaning that the probability of B must be accounted for in the expression. Really, you can think of Bayes' rule in terms of simple algebra on this.

$$P(A|B)P(B) = P(A \cap B) \rightarrow\ P(A|B) = \frac{P(A \cap B)}{P(B)}$$

If A and B are independent, in contrast, one could simply write $$P(A)P(B) = P(A \cap B)$$, because in this case $$P(A|B) = P(A)$$.

The remark about dividing by the number of occurrences of B sounds like perhaps you were thinking of the relative frequency among equally probable outcomes, whereas Bayes' rule applies more generally.

The probability of B here is just 1, is it not?

Why would you think that? $$B$$ is a subset of the outcome space, so has some probability between $$0$$ and $$1$$ (excluding $$0$$; because we don't want to divide by that). $$\mathsf P(B)\in(0..1]$$

Note: $$\mathsf P(B)$$ is not the probability of $$B$$ when given $$B$$. Rather it is the probability of $$B$$ when given no constraints.

If you like: $$\mathsf P(B)=\mathsf P(B\mid\Omega)$$ and so forth, so basically:

$$\mathsf P(A\mid B)=\dfrac{\mathsf P(A\cap B\mid\Omega)}{\mathsf P(B\mid\Omega)}$$

The probability of $$A$$ when given $$B$$ equals the probability of the intersection of $$A$$ and $$B$$ (when given anything), divided by the probability of $$B$$ (when given anything).

• I thought the probability of B is 1 here because we know that B has occurred Jun 24, 2022 at 8:17
• You don't know $B$ has occurred. You are just evaluating the probability that $A$ occurs $\underset{\tiny\text{under the condition that}}{\text{when given}}$ that $B$ occurs. "When given" meaning "should it happen to be the case". Jun 24, 2022 at 8:52
• Or to put it another way: you only treat $B$ as being "given" in measures where you are told to do so. $\mathsf P(A\mid B)$ is the probability for $A$ occurring when $B$ occurs. $\mathsf P(B)$ is the probability for $B$ occurring with no conditions. Jun 24, 2022 at 8:58
• I wonder who can read that tiny text without zooming in $1000\times.$ That final equation is a nice clarifying touch. Jun 25, 2022 at 2:02