Answer
$P(weight~gain~in~excess~and~baby~girl)=0.099495$
Work Step by Step
$20.1$% $=0.201$ and $49.5$% $=0.495$
$P(weight~gain~in~excess)=0.201$
$P(baby~girl)=0.495$
Using the Multiplication Rule for Independent Events (page 282):
$P(weight~gain~in~excess~and~baby~girl)=
P(weight~gain~in~excess)×P(baby~girl)=
0.201×0.495=0.099495$
