Total probability problem I currently have log files all off different sizes, and I am trying to calculate the total probability of an event that occurs being an error. 
For example:

Log file #1: Has a total of 1000 entries and 20 of them entries are errors so the conditional probability for that file would be 20/1000 = 0.02
Log File #2: Has a total of 2500 entries and 35 of them entries are errors so the conditional probability for that file would be 35/2500 = 0.014
Log File #3: Has a total of 5000 entries and 60 of them entries are errors so the conditional probability for that file would be 60/5000 = 0.012

I was wondering how I would get the total probability of an event that occurs being an error for all the log files combined?
Any help would be much appreciated, and if you could use the example I gave to explain to me it would be great.  The events in the log file are completely different, the log files are for completely different systems. 
Would it be (amount of errors/total number of events) = 115/8500 = 0.0135?
 A: From an IT point of view, it doesn't matter that the log files are coming from disparate systems. All of the log events originating from systems under hacker9116's purview constitute a single log stream of events consisting of (State, System) pairs and as a result, the probability of an error can be approximated from the log stream by the law of total probability.
Let $X$ be a discrete random variable taking on Error and Non-error representing the state of the log event. Let $S$ be a discrete random variable taking on non-negative integer values represent the system generating the log event.
Based on the information provided by hacker9116.
$\mathbb{P}(X = \text{Error} | S = s) \approx \begin{cases} \frac{20}{1000} & s = 0 \\ \frac{35}{2500} & s = 1 \\ \frac{60}{5000} & s = 2 \\ 0 & \text{otherwise} \end{cases}$ , $\mathbb{P}(S = s) \approx \begin{cases} \frac{1000}{8500} & s = 0 \\ \frac{2500}{8500} & s = 1 \\ \frac{5000}{8500} & s = 2 \\ 0 & \text{otherwise} \end{cases}$
Given the law of total probability: $\displaystyle \mathbb{P}(X = \text{Error}) = \sum_{s = 0}^{2} \mathbb{P}(X = \text{Error} | S = s) \mathbb{P}(S = s)$,
$\implies \mathbb{P}(X = \text{Error}) = \frac{20}{1000} \frac{1000}{8500} + \frac{35}{2500} \frac{2500}{8500} + \frac{60}{5000}\frac{5000}{8500} = \frac{115}{8500}$.
So I would argue that your approach is correct hacker9116.
