Answer
$\frac{1800}{360360}$
$ \frac{1}{360360}$
Work Step by Step
Given $6W$, $4B$, $5R$, to get a sequence of $BBRWW$, the probabilty would be:
$p_1=(\frac{4}{15})(\frac{3}{14})(\frac{5}{13})(\frac{6}{12})(\frac{5}{11})=\frac{1800}{360360}$
To get a sequence of $2,6,4,9,13$, the probability would be:
$p_2=(\frac{1}{15})(\frac{1}{14})(\frac{1}{13})(\frac{1}{12})(\frac{1}{11})=\frac{1}{360360}$
