Answer
See the picture.
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$
n = 5, p = 0.2 and 1 - p = 0.8
$P(0)={}_5C_{0}\times0.2^0\times0.8^{5}=\frac{5!}{0!\times5!}\times1\times0.32768=1\times1\times0.32768=0.32768$
$P(1)={}_5C_{1}\times0.2^1\times0.8^{4}=\frac{5!}{1!\times4!}\times0.2\times0.8^4=5\times0.2\times0.4096=0.4096$
$P(2)={}_5C_{2}\times0.2^2\times0.8^{3}=\frac{5!}{2!\times3!}\times0.04\times0.512=10\times0.04\times0.512=0.2048$
$P(3)={}_5C_{3}\times0.2^3\times0.8^{2}=\frac{5!}{3!\times2!}\times0.008\times0.64=20\times0.008\times0.64=0.1024$
$P(4)={}_5C_{4}\times0.2^4\times0.8^{1}=\frac{5!}{4!\times1!}\times0.0016\times0.8=5\times0.0016\times0.8=0.0064$
$P(5)={}_5C_{5}\times0.2^5\times0.8^{0}=\frac{5!}{5!\times0!}\times0.00032\times1=1\times0.00032\times1=0.00032$
