# Solving the probability of independent events without the complement

Suppose that virus transmission 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:

$$P(C_1\cap C_2 \cap \cdots\cap C_{500}) = (1 - (\frac{1}{500}))^{500} = 0.37$$

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 independent 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 $(\frac{499}{500})(\frac{499}{500})(\frac{1}{500})(\frac{499}{500}) \cdots (\frac{1}{500})$ as an example of one possible combination.

• Please add the [self-study] tag & read its wiki. – gung - Reinstate Monica May 2 '15 at 16:20
• Your blocked Latex equation is ambiguous because you are missing a close parenthesis. Please add it where you think it should go. – gung - Reinstate Monica May 2 '15 at 16:21

Well, you could use the principle of inclusion-exclusion.

It would not be remotely practical in this instance, but you could use it in principle.

It's a rule for writing the union of $k$ events in terms of intersections of events, by writing it as the singleton events minus the pairwise intersections, plus the triple intersections, and so on to higher orders of intersection with each term alternating.

So for example

\begin{eqnarray} P(A_1\cup A_2\cup A_3) &=& P(A_1)+P(A_2)+P(A_3)\\ &=& -\,[P(A_1\cap A_2)+P(A_1\cap A_3)+P(A_2\cap A_3)]\\ &=& + \, P(A_1\cap A_2\cap A_3). \end{eqnarray}

You could then apply mutual independence to simplify those intersection-events.

This is the idea you were coming to in your edit.