Probability that you get a response to a request for a document I am working on a project where I am running a number of different approaches to request documentation for products. I am able to get the documents from 2 different, independent sources.
When I ask Source 1: I get a valid document 20% of the time
When I ask source 2: I get a valid document 70% of the time.
I want to calculate the overall response rate, is P(1) + P(2) = 90% correct?
REFINING QUESTION BASED ON FEEDBACK:
The following formula is helpful to answer the question: P(A∪B)=P(A)+P(B)−P(A∩B)
Meaning they are dependent events, therefore you would need to subtract the intersection ( when both A and B are true), would it just be the 20% for A? My confusion is if this were rolling dice and P(A) is landing on 3 so1/6 and P(B) is landing on an odd number is 3/6. The intersect is when both A and B land on 3. In this case I don't know when they will respond for the same product so how do I define the intersection ( −P(A∩B)l). That's why originally I misunderstood I can calculate as independent and just add them...
 A: 
For dependent events, you would subtract the intersection ( when both
A and B are true), would it just be the 20% for A?

Not really. The formula is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. So in general it is not true that $P(A\cap B)=P(A)$, if $P(A) \leq  P(B)$. If the events A and B are disjoint, then $P(A\cap B)=0$.

My confusion is if this were rolling dice and P(A) is landing on 3
so 1/6 and P(B) is landing on an odd number is 3/6. The intersect is
when both A and B land on 3

Yes, that is true. So $P(A\cup B)=P(A)+P(B)-P(A)=\frac16+\frac12-\frac16=\frac12$.
Here we have the case where $A$ is a subset of $B$. Thus $P(A\cap B)=P(A)$.

To answer your initial question

I am working on a project where I am running a number of different
approaches to request documentation for products. I am able to get the
documents from 2 different, independent sources.
When I ask Source 1: I get a valid document 20% of the time When I ask
source 2: I get a valid document 70% of the time.

The sources are independent. That means that $P(A\cap B)=P(A)\cdot P(B)$. So the probability that at least one of the two sources send a valid document is
$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)\cdot P(B)$
$=0.2+0.7-0.2\cdot 0.7=0.9-0.14=0.76=76\%$
