In a situation, $n$ samples are drawn, each has a probability of $p$ to evaluate true for a certain criterion. The goal is to find a general formula calculating the probability for "at least $x$ number of the samples to be true."
Work so far:
Probability that exactly $0$ sample is true: $$P(0) = (1-p)^{n}$$
Probability that exactly $1$ sample is true: $$P(1) =(p)^1(1-p)^{n-1}$$
Probability that exactly $a$ samples are true: $$P(a) =\binom{n}a(p)^a(1-p)^{n-a}$$
Probability that at least $2$ samples are true: \begin{align}M(2)&=P(2)+P(3)+\ldots+P(n)\\ &= \sum_{i=2}^{n}{P(i)} \\ &= \sum_{i=2}^{n}{\binom{p}i(p)^i(1-p)^{n-i}} \end{align}
which, leads to the general formula of:
$$M(x) = \sum_{i=x}^{n}{\binom{p}i(p)^i(1-p)^{n-i}}$$
Which also inconveniently seem to have no simplification.
Let $X$ denote the number of the samples that are evaluated to be true out of $n$ samples.
Then $X$ follows binomial distrbuition, that is $X \sim Bin(n,p)$.
$$Pr(X \ge x) = \begin{cases} \sum_{i=x}^n \binom{\color{red}n}{x}p^i (1-p)^{n-i} & ,x \leq n \\ 0&, x > n\end{cases}$$