Question 1021611
Your empirical probability of defective unit is {{{8/40=1/5}}}.
Your empirical probability of non-defective unit is then {{{1-1/5=4/5}}}.
Use the formula,
{{{P(X)=C(20,X)*(1/5)^X*(4/5)^(20-X)}}}
where X is the number of defective units.
a) No defective units {{{X=0}}},
{{{P(0)=C(20,0)*(1/5)^0*(4/5)^(20-0)}}}
{{{P(0)=(4/5)^20}}}
{{{P(0)=1099511627776/95367431640625}}}
{{{P(0)=0.01153}}}
b) 2 defective units, {{{X=2}}},
{{{P(2)=C(20,2)*(1/5)^2*(4/5)^(20-2)}}}
{{{P(2)=190*(1/25)*(68719476736/3814697265625)}}}
{{{P(0)=13056700579840/95367431640625}}}
{{{P(0)=0.13691}}}