Answer
a) $p(X=0)=0.1326$
b) $p(X\leq 2)=0.6658$
c) $p(X\leq 5)=0.981$
Work Step by Step
Let X be the random variable of the number of defective components. then X is distributed as binomial distribution with $n$, $p=0.02$
this means that
$p(X=x)=\frac{n!}{x!(n-x)!}(0.02)^x(0.98)^{n-x},x=0,1,...,n$
a) put n=100 in the previous formula
$p(X=0)=\frac{100!}{x!(100-x)!}(0.02)^1(0.98)^{100-x}=0.1326$
b) put n=102, we can allow 2 defective components
$p(X\leq 2)=p(X=0)+p(X=1)+p(x=2)=0.6658$
c) put n=105, we can allow 5 defective components
$p(X\leq 5)=p(X=0)+p(X=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)=0.981$
