Step-by-step explanation:
[tex]A[/tex]- all of the chosen balls are white [tex]E_{i}-[/tex] result of the die roll is [tex]i, \quad i \in\{1,2,3,4,5,6\}[/tex]
Probabilities:
since the die is fair:
[tex] P\left(E_{i}\right)=\frac{1}{6} \quad \text { for } \quad i \in\{1,2,3,4,5,6\} [/tex]
If the die rolls [tex]i[/tex] we choose a combination of [tex]i[/tex] balls, among 10 black and five white balls, therefore
\[ \begin{array}{c}
P\left(A \mid E_{1}\right)=\frac{\left(\begin{array}{c} 5 \\
1 \end{array}\right)}{\left(\begin{array}{c} 15 \\ 1 \end{array}\right)}=\frac{5}{15}=\frac{1}{3} \\ P\left(A \mid E_{2}\right)=\frac{\left(\begin{array}{c} 5 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 15 \\ 2
\end{array}\right)}=\frac{10}{105}=\frac{2}{21} \\
P\left(A \mid E_{3}\right)=\frac{\left(\begin{array}{c} 5 \\ 3
\end{array}\right)}{\left(\begin{array}{c} 15 \\ 3
\end{array}\right)}=\frac{10}{455}=\frac{2}{91} \\
P\left(A \mid E_{4}\right)=\frac{\left(\begin{array}{c} 5 \\ 4
\end{array}\right)}{\left(\begin{array}{c} 15 \\ 4
\end{array}\right)}=\frac{1}{273} \\
P\left(A \mid E_{5}\right)=\frac{\left(\begin{array}{c} 5 \\ 5
\end{array}\right)}{\left(\begin{array}{c} 15 \\ 5
\end{array}\right)}=\frac{1}{3003} \\
P\left(A \mid E_{6}\right)=\frac{\left(\begin{array}{c} 5 \\ 6
\end{array}\right)}{\left(\begin{array}{c} 15 \\ 6
\end{array}\right)}=0
\end{array} \]
