An urn contains 5 white and 10 black balls. A fair die is rolled and that numberof balls are randomly chosen from the urn. What is the probability that all of theballs selected are white? What is the conditional probability that the die landed on3 if all the balls selected are white?

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} \]

the answer to this question is 3/19

step-by-step explanation: