Question 364984
There are {{{2^4=16}}} possible outcomes.
Count the number of girls for each possible outcome.
{{{BBBB=0}}}
{{{BBBG=1}}}
{{{BBGB=1}}}
{{{BBGG=2}}}
{{{BGBB=1}}}
{{{BGBG=2}}}
{{{BGGB=2}}}
{{{BGGG=3}}}
{{{GBBB=1}}}
{{{GBBG=2}}}
{{{GBGB=2}}}
{{{GBGG=3}}}
{{{GGBB=2}}}
{{{GGBG=3}}}
{{{GGGB=3}}}
{{{GGGG=4}}}
.
.
.
The number of girls range from 0 to 4. 
Count the occurences divided by total number of outcomes to get probability.
{{{X }}},{{{ P(X)}}}
------
{{{0}}},{{{ 1/16}}}
{{{1}}},{{{ 4/16}}}
{{{2}}},{{{ 6/16}}}
{{{3}}},{{{ 4/16}}}
{{{4}}},{{{ 1/16}}}
.
.
.
Mean
{{{mu=sum(X*P(X),X=1,4)=0(1/16)+1(4/16)+2(6/16)+3(4/16)+4(1/16)}}}
{{{mu=sum(X*P(X),X=1,4)=4+12+12+4)/16}}}
{{{mu=sum(X*P(X),X=1,4)=32/16}}}
{{{highlight(mu=sum(X*P(X),X=1,4)=2)}}}
.
.
.
Standard deviation (find variance first)
{{{VAR=sum((X-mu)^2*P(X),X=1,4)=(0-2)^2(1/16)+(1-2)^2(4/16)+(2-2)^2(6/16)+(3-2)^2(4/16)+(4-2)^2(1/16)}}}
{{{VAR=sum((X-mu)^2*P(X),X=1,4)=(4+4+0+4+4)/16}}}
{{{VAR=sum((X-mu)^2*P(X),X=1,4)=12/16}}}
{{{VAR=sum((X-mu)^2*P(X),X=1,4)=3/4}}}
{{{sigma=sqrt(VAR)}}}
{{{highlight(sigma=sqrt(3)/2)}}}