Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:

Given that $P(B) = 0$,

$$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$

Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?

  • 3
    $\begingroup$ you might look at en.wikipedia.org/wiki/Regular_conditional_probability $\endgroup$
    – ShawnD
    Commented Feb 16, 2012 at 19:53
  • $\begingroup$ I would think that $P(A|B)$ would be undefined. However, a quick search of wikipedia shows that there are ways to approximate $B$ by events with nonzero probability and consider a limit. (en.wikipedia.org/wiki/…) $\endgroup$
    – Jeff
    Commented Feb 16, 2012 at 19:56
  • 18
    $\begingroup$ Conditioning on zero-probability events is used quite commonly when one is dealing with continuous random variables, since for a continuous random variable $X$, $P\{X = a\} = 0$ for all $a$, while we still want to talk about $P(B|X=a)$ and even use the law of total probability in the form $$P(B) = \int_{-\infty}^\infty P(B|X=a)f_X(a)\mathrm da.$$ $\endgroup$ Commented Feb 16, 2012 at 19:57

3 Answers 3


The comment by Dilip Sarwate points to conditioning on the level of densities which can be interpreted as conditioning on a family of events of probability zero. It is a probabilistic version of Radon-Nikodym derivative.

One can also condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability. For example, begin with a probability measure $\nu$ on the space $C[0,1]$ of continuous functions such that $\nu(\{f(0)=0\})=0$. Restrict it to the set of functions $f$ such that $|f(0)| \le \epsilon$. Normalize this restriction to a probability measure. If these normalized restrictions converge (in some sense) to a probability measure $\mu$ on $C[0,1]$, we can regard $\mu$ as conditional distribution on the zero-probability event $f(0)=0$.


The definition of conditional probability $$P(A|B)=\frac{P(A \cap B)}{P(B)}$$ means that we can't condition on an event with zero probability; if $P(B)=0$, then $P(A|B)$ is undefined for any event $A$.

Sometimes it makes sense to define conditional probability differently. Let $X$ and $Y$ be jointly continuous random variables; then we can define conditional PDF of $X$ given $Y$ $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$$ for all $y$ such that $f_Y(y)>0$, and using this definition we can define $$P(X\in A|Y=y)=\int_Af_{X|Y}(x|y)dx$$ which implies conditioning on zero probability event $\{Y=y\}$.

The answer to the general OP question is: if $P(B)=0$, the conditional probabilities given event $B$ are undefined, unless a different definition of conditional probability is used.


This argument applies only to probability of events and not for probability with random variables.

Firstly the definition, $$P(A|B)=\dfrac{P(A \cap B)}{P(B)}$$ applies only when $P(B) \neq 0$.

What is it when $P(B)=0$?

Well, $P(A|B)$ is the probability of $A$ happening given that $B$ happened is ridiculous to even ask. How can you be given that an event that cannot happen has happened? So, it shall remain indeterminate in my opinion.

However, some texts, do give that $P(A|B)=0$ and some others assign $1$, or so my teacher has told us that he has seen one book doing the latter (asssigning $1$).

But, it just does not matter.

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    $\begingroup$ There are counterexamples to this argument. Suppose X and Y are independent RVs in [0,1] and we want P(X+Y > 1.5 | X = .8). P(X = .8) = 0, but clearly that probability is P(Y >= .7) = .3. Similarly, P(X+Y > 1.5 | X = .2) is zero. Also, the set where a random variable is valued in a set is an event. (In fact, that's the criteria that defines a random variable -- that it's a measurable function.) $\endgroup$
    – mach
    Commented Feb 15, 2015 at 1:46
  • $\begingroup$ @mach Why .3? Are X and Y uniformly distributed? $\endgroup$
    – BCLC
    Commented May 2, 2015 at 13:22
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    $\begingroup$ Measure zero $\neq$ impossible. $\endgroup$
    – user223391
    Commented Sep 3, 2017 at 13:07

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