Answer
$0.0000000083$
Work Step by Step
To get all the white balls, $p_1=\frac{1}{\ _{53}C_5}$.
To get the red ball, $p_2=\frac{1}{\ _{42}C_1}$.
The end probability is $p=p_1p_2=\frac{1}{\ _{53}C_5}\times\frac{1}{\ _{42}C_1}=\frac{1}{120,526,770}\approx0.0000000083$
