Question

Binomial distribution:
n=19, p=0.07, q= 0.93 x=2 or more

how do I find the probability of x being 2 or more?

Answer 1

Given that \(X\sim\mathcal B(n,p)\), you have
\[\mathbb P(X\ge2)=\sum_{i=2}^n\mathbb P(X=i)=\sum_{i=2}^n\binom nip^{n-i}q^i\]You can do less work by computing the complement probability:
\[\begin{align*}
\mathbb P(X\ge k)&=1-\mathbb P(X<k)\\[1ex]
\implies \mathbb P(X\ge2)&=1-\mathbb P(X<2)\\[1ex]
&=\mathbb P(X=0)+\mathbb P(X=1)
\end{align*}\]where
\[\mathbb P(X=k)=\binom nkp^{n-k}q^k\]