Probability of multiple events occurring concurrently I have two events. Each event will occur once and only once. During each time interval, each event has an independent ~1% chance ($\frac{1017}{120000}$ to be precise) of occurring. How would I calculate the probability of both events occurring concurrently (or after the same number of tests)? Or within one or two tests of each other? Calculating the probability of both occurring during a single interval is easy, but that doesn't consider the probability of one of the events occurring before the other, which intuitively seems like it would have a significant effect.
 A: Let $f_1(k)$ be the probability that the first of your two event occurs for the first-and-only time at interval $k$; and let $p = \frac{1017}{120000}$ be the 'probability of occurring during any time interval'. 
I assume you mean: given that the event has NOT occurred previously to interval $k$, the probability that it occurs during interval $k$ is $p$.
Then $f_1(k) = (1-p)^{k-1}p$. And likewise for the second of your two events, $f_2(k) = (1-p)^{k-1}p$. Let $f(k)$ be the probability of both occurring at interval $k$; then
$$f(k) = f_1(k)f_2(k) = ((1-p)^{k-1}p)^2 = ((1-p)^2)^{(k-1)}p^2$$
So if $f$ equals the total probability of both events occurring 'concurrently' at the same interval $k$, then
$$f = \sum_{k=1}^{\infty} ((1-p)^2)^{(k-1)}p^2$$
$$= p^2 \sum_{k=0}^{\infty} ((1-p)^2)^{k}$$
which, since this is a geometric series with $(1-p)^2<1$, allows us to continue
$$= \frac{p^2}{ 1 - (1-p)^2} = \frac{p^2}{2p-p^2} = \frac{p}{2-p}$$
$$= \frac{1017}{240000 - 1017} = \frac{1017}{238983}$$
A: HINT: if I interpret the problem correctly, the OP asks how to obtain the probability that both independent events occur at least one time after a given number $N$ of equal intervals, knowing the probability $p $ that each occurs in one interval. If so, it is easier to  start from the probability that at least one of the two events does NOT occur after $N$ intervals.
