Answer
$P(1)=\frac{1}{19}=0.05263$
Work Step by Step
$P(x)=({}_{x-1}C_{r-1})p^r(1-p)^{x-r}$
$x=1$, $r=1$, $p=\frac{2}{2+18+18}=\frac{1}{19}$ and $1-p=\frac{18}{19}$
$P(1)=({}_{0}C_{0})(\frac{1}{19})^1(\frac{18}{19})^{0}=1\times\frac{1}{19}\times1=0.05263$
