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 $p=0.5,n=6,k=0$ we have:
$P(0)=\frac{6!}{(6-0)!0!}.(0.5)^0(1-0.5)^{6-0}\approx0.0156$
Substituting $p=0.5,n=6,k=1$ we have:
$P(1)=\frac{6!}{(6-1)!1!}.(0.5)^1(1-0.5)^{6-1}\approx0.0938$
Substituting $p=0.5,n=6,k=2$ we have:
$P(2)=\frac{6!}{(6-2)!2!}.(0.5)^2(1-0.5)^{6-2}\approx0.2344$
Substituting $p=0.5,n=6,k=3$ we have:
$P(3)=\frac{6!}{(6-3)!3!}.(0.5)^3(1-0.5)^{6-3}\approx0.3125$
Substituting $p=0.5,n=6,k=4$ we have:
$P(4)=\frac{6!}{(6-4)!0!}.(0.5)^4(1-0.5)^{6-4}\approx0.2344$
Substituting $p=0.5,n=6,k=5$ we have:
$P(5)=\frac{6!}{(6-5)!5!}.(0.5)^5(1-0.5)^{6-5}\approx0.0938$
Substituting $p=0.5,n=6,k=6$ we have:
$P(6)=\frac{6!}{(6-6)!6!}.(0.5)^6(1-0.5)^{6-6}\approx0.0156$
