Answer
See the picture.
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$
n = 6, p = 0.3 and 1 - p = 0.7
$P(0)={}_6C_{0}\times0.3^0\times0.7^{6}=\frac{6!}{0!\times6!}\times1\times0.7^6=1\times1\times0.7^6=0.1176$
$P(1)={}_6C_{1}\times0.3^1\times0.7^{5}=\frac{6!}{1!\times5!}\times0.3\times0.7^5=6\times0.3\times0.7^5=0.3025$
$P(2)={}_6C_{2}\times0.3^2\times0.7^4=\frac{6!}{2!\times4!}\times0.09\times0.7^4=15\times0.09\times0.7^4=0.3241$
$P(3)={}_6C_{3}\times0.3^3\times0.7^3=\frac{6!}{3!\times3!}\times0.027\times0.343=20\times0.027\times0.343=0.1852$
$P(4)={}_6C_{4}\times0.3^4\times0.7^2=\frac{6!}{4!\times2!}\times0.0081\times0.49=15\times0.0081\times0.49=0.05954$
$P(5)={}_6C_{5}\times0.3^5\times0.7^1=\frac{6!}{5!\times1!}\times0.00243\times0.7=6\times0.00243\times0.7=0.01021$
$P(6)={}_6C_{6}\times0.3^6\times0.7^0=\frac{6!}{6!\times0!}\times0.3^6\times1=1\times0.3^6\times1=0.000729$
