Solving the probability of independent evnts without the complement Suppose that virus transmision in 500 acts of intercourse are mutually independent events and that the probability of transmission in any one act is $\frac{1}{500}$. What is the probability of infection?
So I do know that one way to solve this is to find the probability of complement of the event we are trying to solve. Letting $C_1,C_2,C_3...C_{500}$ denote the events that a virus does not occur during encounters 1,2,....500. The probability of no infection is:
$$P(C_1\cap C_2 \cap....\cap C_{500}) = (1 - (\frac{1}{500}))^{500} = 0.37$$
then to find the probability of infection I would just do : $1 - 0.37 = 0.63$
but my question is how would I find the probability not using the complement? I would have thought since the events are independent and each with probability of $\frac{1}{500}$ that if I multiplied each independet event I could obtain the value, but that is not the case. What am I forgetting to consider if I wanted to calculate this way?  I'm asking more so to have a fuller understanding of both sides of the coin.
Edit: I think I may have figured out what I'm missing in my thinking. In the case of trying to figure out the probability of infection I have to take into account that infection could occur on the first transmission, or the second, or the third,...etc. Also transmission could occur on every interaction or on a few interactions but not all. So in each of these scenarios I would encounter some sort of combination of probabilities like $(\frac{499}{500})(\frac{499}{500})(\frac{1}{500})(\frac{499}{500})......(\frac{1}{500})$ as an example of one possible combination.
 A: The general answer is independent of viruses and intercouses. 
Let $C_1, C_2$ be two events of the same probability $p$ and the question is the probability that at least one of them occurs.
One can say that


*

*$$P(C_1 \cup C_2)=P(C_1)+P(C_2)-P(C_1 \cap C_2).$$
or that

*$$P(C_1\cup C_2)=P\left(\overline{\overline{C_1\cup
    C_2}}\right)=P\left(\overline{ \overline {C_1} \cap \overline{C_2}
    }\right)=1-P\left(\overline {C_1} \cap \overline{C_2}\right).$$


If $C_1$ and $C_2$ are independent then


*

*$$P(C_1 \cup C_2)=P(C_1)+P(C_2)-P(C_1)( C_2)=2p-p^2.$$
or

*$$P(C_1\cup C_2)=1-P\left(\overline
    {C_1}\right)P\left(\overline{C_2}\right)=1-(1-p)^2.$$
Both approaches can be generalized for any number of independent events of the same probability:


*

*$$P(C_1\cup C_2 \cup\cdots \cup C_N)=\sum_{k=1}^N(-1)^{k-1}{n \choose k}p^k$$
or

*$$P(C_1\cup C_2 \cup\cdots \cup C_N)=1-(1-p)^N.$$


The second version is way simpler if the events are independent and of equal probabilities. The first one, however, works in general (not in this special form!).
A: You could compute the geometric series:
$$\frac{1}{500}+\left(\frac{499}{500}\right)\left(\frac{1}{500}\right)+\left(\frac{499}{500}\right)^2\left(\frac{1}{500}\right)+,...,+\left(\frac{499}{500}\right)^{499}\left(\frac{1}{500}\right)\\=\frac{\left(\frac{1}{500}\right)\left(1-\left(\frac{499}{500}\right)^{500}\right)}{1-\left(\frac{499}{500}\right)} $$
This is the sum of the probability of getting infected first time, on the second time up to the last time.
