Answer
$P(between~1~and~4~inclusive)=0.6389$
Work Step by Step
$P(between~1~and~4~inclusive)=P(1\leq X\leq4)=P(1)+P(2)+P(3)+P(4)={}_{20}C_{1}\times0.05^1\times0.95^{19}+{}_{20}C_{2}\times0.05^2\times0.95^{18}+{}_{20}C_{3}\times0.05^3\times0.95^{17}+{}_{20}C_{4}\times0.05^4\times0.95^{16}=\frac{20!}{1!\times19!}\times0.05\times0.95^{19}+\frac{20!}{2!\times18!}\times0.05^2\times0.95^{18}+\frac{20!}{3!\times17!}\times0.05^3\times0.95^{17}+\frac{20!}{4!\times16!}\times0.05^4\times0.95^{16}=0.6389$
