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