Reverse 'At least once' probability I got a little question with probabilities that I can't seem to solve, nor to find a solution anywhere... 
So the thing is, I'm trying to solve the at least once problem, but backwards. 
I know the probability $p_A$ that event $A$ occurs at least once in a repetition of $N$ steps. What is the probability in each step that event $A$ happens? 
 A: If the probability that $A$ occurs in a single step is $q$, and all events are independent, then the probability $p$ that $A$ occurs at least once in $n$ steps is obtained by subtracting the probability that it never occurs from $1$. This gives
$$p=1-(1-q)^n.$$
We can rearrange this to make $q$ the subject:
$$q=1-\sqrt[n]{1-p}.$$
A: \begin{align}
\Pr(A \text{ occurs at least once in $N$ steps}) &= 1- \Pr(A \text{ does not occur in steps $1, 2, ..., N$})\\
&= 1- \prod_{i=1}^N \Pr(A \text{ does not occur in step $i$})\\
&= 1-(1-\Pr(A))^N
\end{align}
Here I have assumed that all steps are independent from each other.
A: If the events have the same probability and they are independent,
\begin{align*}
P(A_1\cup\ldots\cup A_N)&=1-P((A_1\cup\ldots\cup A_N)^c)\\
&=1-P(A_1^c\cap\ldots\cap A_N^c)\\
&=1-P(A_1^c)\ldots P(A_N^c)\\
&=1-P(A_1^c)^N.
\end{align*}
A: If the repetitions are independent
$$P[A\ happens\ at\ least\ once] = 1-P[A\ doesn't\ happen\ at\ any\ step]$$
$$= 1-P[A\ doesn't\ happen\ at\ step\ 1]^N$$
This should give you what you need
