Probability question similar to boy-and-girl problem Problem:
Each of two balls is painted either black or gold, with probability 1/2 in each case, and these events are independent. The two balls are placed in an urn.
a. Given the known information that at least one of the balls is painted gold, because you know the gold paint has been used, compute the conditional probability that both balls are painted gold (as usual, showing the steps of your reasoning).
b. Somebody hands you one of the balls in the urn, and you see that it’s painted gold. What is the probability that both balls are painted gold, and why?

My current thoughts:
So I think I got part A correctly. It is a conditional probability about the likelihood that two gold balls would be taken. I did P(2 Gold Balls)= (1/3)
Part B tells me that I'm given one of the balls as Gold. It asks me what are the chances that both balls are painted gold. It can't just be the probability of one of the gold balls (1/2). How do I deal with this question? :( Does any one know what this actually means?!?!
Correct me if I'm wrong but both part A and B are clearly different questions related to probability. Why are they different? I really appreciate your feedback and help.
 A: Suppose further that the balls are labeled with a $1$ and a $2$.  This will help us organize our thoughts and create an equiprobable sample space.
For the first part, it is not $Pr(\text{both gold})$, it is $Pr(\text{both gold} \color{red}{| \text{at least one gold}})$.  This is a very important distinction to make.
Our original sample space where we don't care about what order we draw balls in is:  $\{G_1G_2, G_1B_2, B_1G_2, B_1B_2\}$.  Since we are working with the conditional probability that at least one of the balls is gold, we can remove $B_1B_2$ from our sample space.  Our conditioned sample space is then $\{G_1G_2,G_1B_2,B_1G_2\}$, each of which are equally likely to occur.  The probability that both are gold given that we know at least one of the balls is gold is then $\frac{1}{3}$.
(compare this to the boy-girl paradox where we are told at least one of the two children is a girl)
Our next part we will need to worry about which order the balls are drawn as well, so we put a bit more effort into our sample space.
Here, it is $Pr(\text{both gold}\color{red}{|\text{first ball drawn is gold}})$
Our sample space originally where order matters is $\{G_1G_2,G_2G_1,G_1B_2,B_2G_1,B_1G_2,G_2B_1,B_1B_2,B_2B_1\}$.  Since we are conditioning on the first ball seen being gold, we may remove any and all outcomes from our sample space where the first ball was black.

  This leaves us with the conditioned sample space $\{G_1G_2,G_2G_1,G_1B_2,G_2B_1\}$, again with each equally likely to occur.  Exactly two of the four outcomes here result in both being gold, so the probability is $\frac{2}{4}=\frac{1}{2}$.

(compare this to the boy-girl paradox where we are told the eldest child is a girl)
I will leave the question of why they are different questions to you, but it has something to do with the amount of information we are given.
