1) Binomial distribution

Given $P=0.01\\q=1-p=0.99\\n=10$

$P(X=r)= nC_r \ p^r \ q^{n-r}$

$for \ P(X \gt1)=1-P(X\le1)$

i.e $1-P(X=0)-P(X=1)$

=$1-10 C_0 (0.01)^0(0.99)^{10-0}-10C_1(0.01)^1(0.99)^{10-1}$

= $1-10 C_0 (1)(0.99)^{10}-10C_1(0.01)^1(0.99)^9$

=$1-0.9044-0.09135$

$P(X\gt1)=0.00425$

2) Poison distribution

given ,m=n.p

m=10(0.01)

m=0.1

P(X=r)=$e^{-m} \frac{m^r}{r!}$

$P(X\gt 1)=1-P(X\lt 1)$

=$1-P(X=0)-P(X=1)$

=$1-e^{-0.1} \frac {(0.1)^0}{0!}-e^{-0.1} \frac{(0.1)^1}{1!} $

=$1-0.9048-0.0905$

$P(X\gt 1) =0.0047$