# Probability of A given B or C

I know some formulae related to conditional probabilities of events conditional on intersection of two events.

But I have been unable to find any formula for the case where the condition is union of two or more events.

That is, how can I find

$$P(A | B\text{ or }C)$$

And in general

$$P(A | B\text{ or }C\text{ or }\dots\text{ or }X)$$

And what can be the continuous analog?

$$p(x | y \in [a,b])$$

When I already know $$p(x | y=t), \forall t \in [a,b]$$

Edit: An approach I tried

$$P(A |B\text{ or }C) = \frac{P ((A\text{ and }B)\text{ or }(A\text{ and }C))}{P(B\text{ or }C)}$$

$$=\frac {P(A,B)+P(A,C)-P(A,B,C)} {P(B\text{ or }C)}$$

For the continuous case I asked, I assume that $$y$$ can take only one value at a time, so the $$P(A,B,C)$$ term can be ignored.

I have not tried a derivation, but just by analogy, I guess the continuous expression would be

$$p(x|y \in [a,b]) = \frac { \int_a^b p(x|y) p(y) dy} {P(y \in [a,b])}$$

Is the approach for the discrete case correct?

I have just used analogy for the continuous case, is the formula I guessed correct? If yes, how to prove that? If not, what is the correct formula?

• just define the event $D = B$ union $C$ and then calculate $P(A|D )$. Commented Apr 19, 2020 at 3:48
• I have edited the question and added an approach I tried, please comment on the continuous part. And I am sorry, I tried several times but I was unable to fix the formatting error. Commented Apr 19, 2020 at 16:47
• Isn't this just a logical follow on from some simpler rules? As in the probability of B union C is P(B) + P(C) - P(B intersection C), and for a sequence of events, that is the union of this result and the next possible event, applied as many times as necessary. Commented Apr 26, 2020 at 21:38

1) $$P(A | B \text{ or } C)=P(A|B\cup C)=\frac{P(A\cap(B\cup C))}{P(B\cup C)}$$

2)$$P(A | B \text{ or } C \text{ or } \dots \color{red}{\text{or }X})$$

what is $$B \text{ or } X$$? If $$X$$ is a random variable, I think it is only valid if we use it like $$B\cup \{X\in E\}=\{\omega \in \Omega \mid \omega \in B \text{ or } x(\omega)\in E\}$$.

so $$P(A | B \text{ or } C \text{ or } \cdots \text{ or } \{X \in E\})$$ can be easily calculated by defining $$D=B \cup C \cup \cdots \cup \{X \in E\}$$.

3) $$P(X= x|Y\in [a,b])$$ for the case $$Y$$ is a continues random variable You can easily calculate it if you knowing $$P(X\leq x|Y\in [a,b])$$.

$$P(X\leq x|Y\in [a,b])=P(\{X \leq x\}|\{Y\in [a,b]\})=\frac{P(\{X \leq x\} \cap \{Y\in [a,b]\})}{P(\{Y\in [a,b]\})}=\frac{\int_{-\infty}^{x}\int_{y\in [a,b]}f_{(X,Y)}(t , y) dy dt}{P(\{Y\in [a,b]\})}=\frac{\int_{-\infty}^{x} \int_{y\in [a,b]}p(t | y)p(y) dy dt}{P(\{Y\in [a,b]\})}$$.

• Thank you for the answer. Commented Apr 27, 2020 at 16:27
• You are welcome! Commented Apr 27, 2020 at 16:29
• And regarding the second point you raised, I am sorry if I used confusing (or wrong) notation, by X, I simply meant another event and not a random variable when I wrote $P(A|B or C or ..... or X)$, just like B and C, and not a random variable. Commented Apr 27, 2020 at 16:30