probability
Say we have $P(A) = 0.60,\, P(B) = 0.50$ .
Normally to find the probability of at least one happening, we find the probability of neither of them happening:
$$P(A^c \cap B^c) = 0.40 \times 0.50 = 0.20$$
and then subtract it from $1$ ( getting $0.80$). I know that this is incorrect because we are also given that $P(A \cap B) = 0.35$, thus $A,B$ are not independent. How would one go about finding the probability of at least one occurring in this case?
Whenever you need to find the probability of at least one thing happening, you can instead ask "What is the probability that none of them happen?" and subtract from $1$ (since the complementary event to "none happen" is "at least one happens").
For each of the individual events, we find the probability it does not happen by subtracting the probability that it does happen from 1. We have
Since the events are independent, the probability no event happens is the product of the individual probabilities, which is $133/1000$.
Based on the calculation above $$ \Pr(\text{at least one event}) = 1 - \Pr(\text{none of the events}) = 1 - \frac{133}{1000} = \frac{867}{1000} = 86.7\%. $$
Hint: Add up the area you want to calculate! 
Best Answer
The event $C=$"at least one from A or B happened" is the same as the event $C=A\cup B$.
So.. just use one of the most basic results in probabilities: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.6-0.35=0.75.$$