# Link between P(A) and P(A|B)

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 ?
• 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) Commented Sep 9, 2017 at 9:13
• P(A, B) = P(A) * P(B) only when A, B are independent. Commented Sep 9, 2017 at 9:14
• Yes, you are right, I just assumed that for some reason. Commented Sep 9, 2017 at 9:15
• – Tim
Commented Sep 11, 2017 at 8:54

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

example:

• 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.

• 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? Commented Sep 9, 2017 at 9:15
• 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. Commented Sep 9, 2017 at 9:19
• 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. Commented Sep 9, 2017 at 9:24
• 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. Commented Sep 11, 2017 at 8:43
• Yes, I should have used the terms: 1) event happens almost never 2) event happens almost surely Commented 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 $$1-\dfrac{1-\mathbb P (A)}{\mathbb P (B)} \le \mathbb P (A\mid B) \le \dfrac{\mathbb P (A)}{\mathbb P (B)}$$

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