How do I add multiple non-exclusive probabilities

Question

What I've learned about adding probabilities is that mutually exclusive probabilities you can just add together: $P(A \ or \ B)=P(A)+P(B)$. If they are non-mutually exclusive the probability $A$ or $B$ will occur is: $P(A \ or \ B)=P(A)+P(B)-P(A \ and \ B)$ where the last term is the overlap of the two events. When it comes to dealing with two probabilities I've been able to solve, but I have a problem where I have 5 different probabilities that an a certain amount of cars will go through an intersection within the time span of 5 minutes. The following table has the probabilities I am working with:

I need to find out the probability at least one car will pass in the time span. I thought the way to do this would be to add up the probabilities from 1 cars to 5, then subtract them from one another due to the overlap of the events because these are non-mutually exclusive. $$(0.28+0.15+0.10+0.07+0.04)-(0.28-0.15-0.10-0.07-0.04)=0.72$$ This looked OK to me because the number is less than $1$, which it must be less than, but apparently this isn't the right answer.

What is the right way to add more than two probabilities together?

Answer 1

If you count all the probabilities from your table, they will add up to 1.

The event that (exactly) 5 cars pass through the intersection in a span of 5 minutes, cannot overlap the event that (exactly) 1 car pass through the intersection in the same span of 5 minutes.

Illustration :

This may help you understand, these probabilities could arise from a research with 1000 data. Each data is a number of total cars that pass through intersection at different span of 5 minutes. Probability only 1 car that pass through, is the ratio of the total number of data of 1 car pass through the intersection, 280, and the total number of data, 1000. Overall, the each proportion makes up your probability table. These events are mutually exclusive.

The probability of at least one car pass through the intersection can be counted in these ways, with $C=c$ as the event of total of $c$ cars pass through intersection in a span of 5 minutes :

1. $$ P(C>0)= \sum_{i=1}^{5} P(C=i) $$
2. $$ P(C>0)= 1 - P(C=0) $$

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An example where two events are not mutually exclusive are $C>1$ and $C>2$. (Recall the illustration above) This is because all the data that satisfy $C>2$ will also satisfy $C>1$, so there are overlaps.