By the binomial distribution, this probability is
P(X<3)=P(X=0)+P(X=1)+P(X=2)=(200)(0.05)0(0.95)20−0+(201)(0.05)1(0.95)20−1+(202)(0.05)2(0.95)20−2=(0.95)18(0.952+20(0.05)(0.95)+190(0.05)2)=0.9245P(X<3)=P(X=0)+P(X=1)+P(X=2)=\binom{20}{0}(0.05)^0(0.95)^{20-0}+\binom{20}{1}(0.05)^1(0.95)^{20-1}+\binom{20}{2}(0.05)^2(0.95)^{20-2}=(0.95)^{18} (0.95^2 +20(0.05)(0.95)+190(0.05)^2 )=0.9245P(X<3)=P(X=0)+P(X=1)+P(X=2)=(020)(0.05)0(0.95)20−0+(120)(0.05)1(0.95)20−1+(220)(0.05)2(0.95)20−2=(0.95)18(0.952+20(0.05)(0.95)+190(0.05)2)=0.9245
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