Question 415070
The pmf is {{{p(x) = 3Cx*(1/2)^(3-x)*(1/2)^x = 3Cx/8}}}.  Hence

X ----> 0        1        2       3
p(x)--> 0.125    0.375    0.375   0.125

==> {{{mu = 0.375 + 2*0.375 + 3*0.125 = 1.5}}}
==> {{{Var(X) = E(X^2) - (E(X))^2 = 3 - 1.5^2 = 0.75}}}.

A shorter solution is as follows:  The mean {{{mu}}} for a binomial experiment is n*p = 3*0.5 = 1.5, while the variance is n*p*q = 3*0.5*0.5 = 0.75.  (Here, n = 3, and p = q = 0.5.)