\(\mathbb P(9\le X\le 10)=\mathbb P(X=9)+\mathbb P(X=10)\)
Since
\(\mathbb P(X=x;n,p)=\dbinom nxp^x(1-p)^{n-x}\)
you have
\(\mathbb P(9\le X\le10)=\dbinom{20}90.4^9(1-0.4)^{20-9}+\dbinom{20}{10}0.4^{10}(1-0.4)^{20-10}\)
\(\mathbb P(9\le X\le10)=\dbinom{20}90.4^90.6^{11}+\dbinom{20}{10}0.4^{10}0.6^{10}\)
\(\mathbb P(9\le X\le10)\approx0.2769\)
