A bag contains one marble which is either green or blue, with equal
probabilities. A green marble is put in the bag (so there are 2
marbles now), and then a random marble is taken out. The marble taken
out is green. What is the probability that the remaining marble is
also green?
Source: Stat 110, Joe Blitzstein
[The solution by the instructor uses the Law of Total Probability but I do not intend to use that.]
Attempt: Denote by $XY$ marble of colour $X$ remains in the bag and the marble of colour $Y$ is taken out. The possibilities are $GG, BG$ and $GB$ out of which the first two are events that we consider. This gives us $P = \frac{2}{3}$ which matches the answer.
My doubt is $GG$ occurs with twice the probability of that of $BG$ (or $GB$) since when both are green, you either take out the marble that you introduced later or you take out the one that's lying inside since the beginning of the experiment. How is that not influencing the result?
Edit: While writing this, I forgot both marbles are supposed to be green and not that just a green marble is to be taken out. So in that case, Favourable : $\{G_1G_2,G_2G_1\}$, Possible – $\{G_1G_2,G_2G_1,BG\}$.
Best Answer
Your attempt has two mistakes, but coincidentally arrives at the right answer.
Mistake 1: the three outcomes $GG$, $BG$, $GB$ are not equally probable. The first outcome has probability $1/2$ (as long as the original marble is green, this will happen). The other two outcomes happen with probability $1/4$ each.
Mistake 2: The problem is really asking for a conditional probability (namely, probability of the remaining marble being green conditioning on the event that you drew a green marble). You only computed the probability that the other marble is green (using the incorrect probabilities from Mistake 1), and ignored the fact that you drew a green marble as well. Using the correct probabilities above, we have $\frac{P(GG)}{P(GG) + P(BG)} = \frac{1/2}{1/2 + 1/4} = \frac{2}{3}$.