Answer
$P(3~or~fewer)=0.9841$
Work Step by Step
$P(3~or~fewer)=P(0)+P(1)+P(2)+P(3)={}_{20}C_{0}\times0.05^0\times0.95^{20}+{}_{20}C_{1}\times0.05^1\times0.95^{19}+{}_{20}C_{2}\times0.05^2\times0.95^{18}+{}_{20}C_{3}\times0.05^3\times0.95^{17}=\frac{20!}{0!\times20!}\times1\times0.95^{20}+\frac{20!}{1!\times19!}\times0.05\times0.95^{19}+\frac{20!}{2!\times18!}\times0.05^2\times0.95^{18}+\frac{20!}{3!\times17!}\times0.05^3\times0.95^{17}=0.9841$
