I feel like there's no link between P(A) and P(A|B), where A and B are random events. For example, P(A) can be 0.3 and P(A|B) can be 0.9.

However, I'm not so sure of this when P(A) is either 0 or 1. I mean:

  1. P(A)=0 => P(A|B)=0, for every random event B ?
  2. P(A)=1 => P(A|B)=1, for every random event B ?
  • $\begingroup$ P(A|B) = P(A,B) / P(B), P(A,B) = P(A) * P(B), therefore, you are right that if P(A) = 0, P(A|B) will always be 0, regardless of P(B) $\endgroup$
    – yasar
    Sep 9, 2017 at 9:13
  • 2
    $\begingroup$ P(A, B) = P(A) * P(B) only when A, B are independent. $\endgroup$
    – user_anon
    Sep 9, 2017 at 9:14
  • $\begingroup$ Yes, you are right, I just assumed that for some reason. $\endgroup$
    – yasar
    Sep 9, 2017 at 9:15
  • $\begingroup$ Check also stats.stackexchange.com/questions/246009/… $\endgroup$
    – Tim
    Sep 11, 2017 at 8:54

2 Answers 2


P(A) and P(A|B) are linked, since the occurrence of B does affect the probability of the occurrence of A.


  • P(A) = You pass your exam = 0.3
  • P(B) = You study for you exam
  • P(A|B) = You studied, so there is a higher chance of passing your exam = 0.9

However, point 1 and 2 are true:

1) P(A) = 0, event that will not occur. No prior event can change that.

2) P(A) = 1, event that will definitely occur. No prior event can change that.

  • $\begingroup$ I thought about the link between the Values of P(A) and P(A|B). P(A) can be every number in (0,1) and P(A|B) any other number in (0,1)... Why 1) and 2) are true? $\endgroup$
    – user_anon
    Sep 9, 2017 at 9:15
  • $\begingroup$ OK. So, if you have only the value of P(A) and have to calculate P(A|B), then you can "calculate" it only if P(A) is 0 or 1, right? If P(A) is in the interval (0,1) then you need extra info to calculate P(A|B), I guess. $\endgroup$
    – user_anon
    Sep 9, 2017 at 9:19
  • $\begingroup$ In theory you are assuming P(A) ≠ 0 and P(A) ≠ 1, that is trivial. Yes, the extra info is the dependence of two events. $\endgroup$ Sep 9, 2017 at 9:24
  • 3
    $\begingroup$ A more precise version of (1) and (2) would be true, but as written, they are not quite right (in the general case). That the probability measure of an event is 1 does not imply the event "will definitely occur." A zero probability does not imply the event will not occur. Example: Let $X$ be a random variable that follows the standard normal distribution. $P(X=0) = 0$ but that event can occur. $\endgroup$ Sep 11, 2017 at 8:43
  • $\begingroup$ Yes, I should have used the terms: 1) event happens almost never 2) event happens almost surely $\endgroup$ Sep 11, 2017 at 8:51

You have issues when $\mathbb P (B)=0$ but is not impossible

Consider for example $X$ a random variable with a standard Gaussian distribution $\mathcal N(0,1)$

  1. Let $A$ be the event $X^2=4$ and $B$ the event $X=-2$. Then $\mathbb P (A)=0$ but $\mathbb P (A\mid B)=1$

  2. Let $A$ be the event $X^2 \not = 4$ and $B$ the event $X=-2$. Then $\mathbb P (A)=1$ but $\mathbb P (A\mid B)=0$

If you know $\mathbb P (B)$, then there are bounds relating $\mathbb P (A)$ and $\mathbb P (A\mid B)$: $$\mathbb P (A\mid B){\mathbb P (B)} \le \mathbb P (A) \le \mathbb P (A\mid B)\mathbb P (B)+(1-\mathbb P (B))$$ which if $\mathbb P (B)$ is non zero is equivalent to $$1-\dfrac{1-\mathbb P (A)}{\mathbb P (B)} \le \mathbb P (A\mid B) \le \dfrac{\mathbb P (A)}{\mathbb P (B)}$$

  • $\begingroup$ Thanks. Do those inequalities have a well-known name? $\endgroup$
    – user_anon
    Sep 11, 2017 at 10:02
  • 1
    $\begingroup$ @user_anon - Perhaps somebody else has given them a name. I had never seen them before trying to answer your question $\endgroup$
    – Henry
    Sep 11, 2017 at 11:59

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