1. ## Fun probability question

A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability is w/w+b+n.

2. Originally Posted by vexiked
A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability is w/w+b+n.
Again, I suspect no one is quite sure what this question means.
Read this. “Show that the conditional probability is w/w+b+n”
Does that make any sense to you?
“the conditional probability” of what?

3. The problem should state:

A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability that the first ball selected was white is w/w+b+n.

4. Originally Posted by vexiked
A bowl contains w white balls and b black balls. One ball is selected at random from the bowl, its color noted, and is returned to the bowl along with n additonal balls of the same color. Another ball is randomly selected (now containing w + b + n balls) and is observed black. Show that the conditional probability that the first ball selected was white is w/w+b+n.
You are asked to find $\displaystyle P(W_1|B_2)$ and that is equal to $\displaystyle \frac{{P(B_2 |W_1 )P(W_1 )}}{{P(B_2 |W_1 )P(W_1 ) + P(B_2 |B_1 )P(B_1 )}}$.

I will do part of it for you. $\displaystyle P(B_2 |B_1 )P(B_1 )=\frac{(n+b)}{w+(n+b)}\cdot\frac{b}{w+b}$

5. Thank you! This makes things alot more clear for me