Question: On average 15% of the population have negative side effects on a particular drug A. 21 patients are chosen randomly. What is the probability that at least 2 of them have negative side effects.
Solution: Let $X$ be the number of patients who have negative side effects, then $X \sim Bin(n,p)$. Furthermore we know that $\mathbb{E}[X] = np = 0.15 \Rightarrow p = \frac{0.15}{n}$ Here $n=21$ thus: $p = \frac{0.15}{21}$. The probability that at least 2 have negative side effects is:
$$ \mathbb{P}(X \geq 2) = 1 - \mathbb{P}(X < 2) = 1 - \mathbb{P}(X = 0) -\mathbb{P}(X = 1) $$
$$ \mathbb{P}(X = 0) = {21\choose 0} p^0(1-p)^{21} \approx 0.86 $$
$$ \mathbb{P}(X = 1) = {21\choose 1} p^1(1-p)^{20} \approx 0.12 $$
Thus $$ \mathbb{P}(X \geq 2) \approx 1 - 0.86 - 0.12 $$
$p = 0.15, (1-p) = 0.85.$
So, the computation is
$$1 ~~- $$
$$\left\{ ~\left[ ~\binom{21}{0} \times (0.15)^0 \times (0.85)^{(21)} ~\right] ~+~ ~\left[ ~\binom{21}{1} \times (0.15)^1 \times (0.85)^{(20)} ~\right] ~\right\}.$$