Answer
See the picture.
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$
n = 10, p = 0.2 and 1 - p = 0.8
$P(0)={}_{10}C_{0}\times0.2^0\times0.8^{10}=\frac{10!}{0!\times10!}\times1\times0.8^{10}=1\times1\times0.8^{10}=0.1074$
$P(1)={}_{10}C_{1}\times0.2^1\times0.8^9=\frac{10!}{1!\times9!}\times0.2\times\times0.8^9=10\times0.2\times0.8^9=0.2684$
$P(2)={}_{10}C_{2}\times0.2^2\times0.8^8=\frac{10!}{2!\times8!}\times0.2^2\times0.8^8=45\times0.2^2\times0.8^8=0.3020$
$P(3)={}_{10}C_{3}\times0.2^3\times0.8^7=\frac{10!}{3!\times7!}\times0.2^3\times0.8^7=120\times0.2^3\times0.8^7=0.2013$
$P(4)={}_{10}C_{4}\times0.2^4\times0.8^6=\frac{10!}{4!\times6!}\times0.2^4\times0.8^6=210\times0.2^4\times0.8^6=0.08808$
$P(5)={}_{10}C_{5}\times0.2^5\times0.8^5=\frac{10!}{5!\times5!}\times0.2^5\times0.8^5=252\times0.2^5\times0.8^5=0.02642$
$P(6)={}_{10}C_{6}\times0.2^6\times0.8^4=\frac{10!}{6!\times4!}\times0.2^6\times0.8^4=210\times0.2^6\times0.8^4=0.005505$
$P(7)={}_{10}C_{7}\times0.2^7\times0.8^3=\frac{10!}{7!\times3!}\times0.2^7\times0.8^3=120\times0.2^7\times0.8^3=0.0007864$
$P(8)={}_{10}C_{8}\times0.2^8\times0.8^2=\frac{10!}{8!\times2!}\times0.2^8\times0.8^2=45\times0.2^8\times0.8^2=0.00007373$
$P(9)={}_{10}C_{9}\times0.2^9\times0.8^1=\frac{10!}{9!\times1!}\times0.2^9\times0.8^1=10\times0.2^9\times0.8^1=0.000004096$
$P(10)={}_{10}C_{10}\times0.2^{10}\times0.8^0=\frac{10!}{10!\times0!}\times0.2^{10}\times1=1\times0.2^{10}=0.0000001024$
