Question

Assume that you have 5 red balls and 9 blue balls from which randomly to select two without replacement.

What is the probability that exactly one blue ball is chosen?

Answer 1

In total you have 9+5=14 balls. Out of those 14 balls, 9 are blue, so when you select a blue one you have a probability of 9/14 = 0.6428 = 64.24 % chance to select a blue ball. The second time you have removed one blue ball from the bag, so now you have 13 balls and 8 blue ones. Therefore you have 8/13=0.6153 = 61.53% to select a blue ball again.

I hope this answers your question because I wasn't entirely sure if this is what you were asking.

Answer 2

This is solved using the hypergeometric distribution;
\[P(1blue)=\frac{\left(\begin{matrix}9 \\ 1\end{matrix}\right)\left(\begin{matrix}5 \\ 1\end{matrix}\right)}{\left(\begin{matrix}14 \\ 2\end{matrix}\right)}=\frac{9\times5 \times2}{14\times 13}=?\]