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The probability of an independent event happening to a thing is $0 \leq p \leq 1$. There are $0 \leq n$ discrete things. What is the probability $P$ that any combination of $0 \leq k \leq n$ do not experience the event?

I can work this out by hand for any two of three things ($n = 3, k = 2$). Let's say the three things are $A, B,$ and $C:$

| Good scenario     | Probability     |
|-------------------|-----------------|
| A and B and not C | p * p * (1 - p) |
| A and C and not B | p * (1 - p) * p |
| B and C and not A | (1 - p) * p * p |

So $P(\text{Good scenario}) = p^2 \cdot (1-p)$, and there are 3 possible good scenarios.

So what's the probability of any of these scenarios happening? I think it's the probability of not(every good scenario fails)... $1 - (1 - P(\text{Good scenario}))^3$

But how do I generalize this? Is the following correct?

$$ P(\text{Good scenario}) = p^k \cdot (1 - p)^{n - k} $$ $$ P = 1 - (1 - P(\text{Good scenario}))^{n \choose k} $$

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  • $\begingroup$ For any two of three things, you have successfully shown that each one individually happens with probability $p^2 (1-p)$ but because all of these consist of a "Good scenario" we must add these together to get a probability of a Good scenario to be $3p^2 (1-p)$ $\endgroup$ – WaveX Dec 31 '19 at 16:58
  • $\begingroup$ The phrasing is not clear. Do you mean "having specified a set of $k$ things, what is the probability that those $k$ $\textit {and no other}$ things do not experience the event?" Or do you mean "what is the probability that the set of things which do not experience the event has exactly $k$ elements?" or do you mean something else? $\endgroup$ – lulu Dec 31 '19 at 17:03
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This is exactly the definition of a binomial distribution, whose probability mass function is

$$\mathrm{Pr}[k\textrm{ out of }n\textrm{ objects experience a probability-}p\textrm{ event}]\equiv \mathrm{Pr}(k; n, p)=\binom{n}{k}p^k(1-p)^{n-k}$$

where $\binom{n}{k}$ is the number of ways to choose $k$ out of the $n$ objects (i.e., the number of "good scenarios", and $p^k(1-p)^{n-k}$ is the probability of each one of these "good scenarios" happening.

This is the same as you have calculated. The final step you were missing is that all the "good scenarios" are pairwise disjoint events, so the probability of their union is just the sum of their probabilities.

Note: the title of your question says "Probability that any $k$ of $n$ do not get an event", but in your example you calculate the probability that any $k$ of $n$ do get the event. It is not clear to me which one you are looking for. My answer gives the probability of getting the event: if you want the probability of not getting it, just replace $p$ by $1-p$.

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