Answer
See below
Work Step by Step
The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$
Substituting $p=0.16,n=4,k=0$ we have:
$P(0)=\frac{4!}{(4-0)!0!}.(0.16)^0(1-0.16)^{4-0}\approx0.498$
Substituting $p=0.16,n=4,k=1$ we have:
$P(1)=\frac{4!}{(4-1)!1!}.(0.16)^1(1-0.16)^{4-1}\approx0.379$
Substituting $p=0.16,n=4,k=2$ we have:
$P(2)=\frac{4!}{(4-2)!2!}.(0.16)^2(1-0.16)^{4-2}\approx0.108$
Substituting $p=0.16,n=4,k=3$ we have:
$P(3)=\frac{4!}{(4-3)!3!}.(0.16)^3(1-0.16)^{4-3}\approx0.01376$
Substituting $p=0.16,n=4,k=4$ we have:
$P(4)=\frac{4!}{(4-4)!4!}.(0.16)^4(1-0.16)^{4-4}\approx0.000655$
