I know the probability of an event to occur, so how do I find how many times the event must be attempted to get a reasonably chance of success? If there is a 2% chance of event X to occur, how many times must X be attempted to have a reasonable probability of occurring at least once?
What is the formula used called, and how do I solve this?
 A: As @Greenparker mentioned above, the number of successes (success defined as event X occurs) out of n trials follows a Binomial distribution: $Y \sim Bin(n, 0.02)$
You just need to find $P(Y \geq 1) \geq a$ , where $a$ is the probability of event occurring at least once, the above probability is equivalent to
$$ 1 - P(Y = 0) \geq a $$
$$ 1 - \binom{n}{0}(1 - .02)^n \geq a $$
which simplifies to 
$$ (1 - .02)^n \leq 1 - a $$
taking log of both sides, and dividing to $log(1-.02) $: 
$$n \geq \frac{log(1-a)}{log(1-.02)}$$
If we want .9 probability of success then n must be at least 114
for $a = .8$: $n >= 80$
for $a = .7$: $n >= 60$
A: First lets find the probability of seeing the even occur at least once in $n$ attempts. Let $Y$ be the number of times the event occurs out of $n$ attempts, where $X$ is the event, and $P(X$ occurs$)$ = .2
Then assuming the attempts are independent, $Y \sim Binomial(0.2, n)$. 
Now essentially, $P(Y \geq 1) = 1  - P(Y = 0)$.
$P(Y = 0)$ is the probability that $X$ never occurs in all $n$ draws, you can find this.
Now to answer your question, if you define reasonable probability as some number $a$, you need to find $n$ such that $P(Y \geq 1) \geq a$.
A: My intuitive answer is 50 times, because you have a 1 in 50 chance (2%).
So, P(X) = 0.02. but that's the probability you get it to happen at least once. You can lower it to, say, 25 times, but then the chance it happens is lower.
