Answer
a) value of X that most likely is 5 because it has the highest probability
b) values of X that least likely are 0 and 10 because they have the least probability
Work Step by Step
Given that, n=10, p=0.5, then,
$p(X=x)=\frac{10!}{x!(10-x)!}(0.5)^{x}(0.5)^{n-x}, x=0,1,...,10$
$p(X=0)=\frac{10!}{0!(10)!}(0.5)^{0}(0.5)^{10}\approx 0.001$
$p(X=1)=\frac{10!}{1!(9)!}(0.5)^{1}(0.5)^{9}\approx 0.0098$
$p(X=2)=\frac{10!}{2!(8)!}(0.5)^{2}(0.5)^{8}\approx 0.0439$
$p(X=3)=\frac{10!}{3!(7)!}(0.5)^{3}(0.5)^{7}\approx 0.1172$
$p(X=4)=\frac{10!}{4!(6)!}(0.5)^{4}(0.5)^{6}\approx 0.2051$
$p(X=5)=\frac{10!}{5!(5)!}(0.5)^{5}(0.5)^{5}\approx 0.2461$
$p(X=6)=\frac{10!}{6!(4)!}(0.5)^{6}(0.5)^{4}\approx 0.2051$
$p(X=7)=\frac{10!}{7!(3)!}(0.5)^{7}(0.5)^{3}\approx 0.1172$
$p(X=8)=\frac{10!}{8!(2)!}(0.5)^{8}(0.5)^{2}\approx 0.0439$
$p(X=9)=\frac{10!}{9!(1)!}(0.5)^{9}(0.5)^{1}\approx 0.0098$
$p(X=10)=\frac{0!}{10!(10)!}(0.5)^{10}(0.5)^{0}\approx 0.001$
a) value of X that most likely is 5 because it has the highest probability
b) values of X that least likely are 0 and 10 because they have the least probability
