Draw a green marble from a bag that already contains a green or a blue marble.

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 $\frac{P(GG)}{P(GG) + P(BG)} = \frac{1/2}{1/2 + 1/4} = \frac{2}{3}$.

You have drawn a green marble. If the remaining marble is also green, it implies that the bag originally had a green marble. So we want to compute the conditional probability: p = P(original marble was green | green marble was drawn).

Let's use Bayes's rule: $P(A|B) = \frac{P(AB)}{P(B)} = \frac{P(B|A)P(A)}{P(AB) + P(A^cB)}$

where $A$ and $B$ are some events and $A^c$ is the complement of $A$.

Let's define: $A$ = original marble was green ($A^c$ = original marble was blue),

$B$ = green marble was drawn,

$P(A) = 1/2$, as given in the problem.

Then we deduce that:

$P(A^c) = 1 - P(A) = 1/2$.

$P(B|A) = 1$, since in this case the bag will contain 2 green marbles.

$P(B|A^c) = 1/2$, since in this case the bag contains 1 green and 1 blue marble.

$P(AB) = P(B|A)P(A) = 1 \times 1/2 = 1/2$

$P(A^cB) = P(B|A^c)P(A^c) = 1/2 \times 1/2 = 1/4$

Substituting in our formula for $P(A|B)$, we get: $P(A|B) = \dfrac{1 \times 1/2}{1/2 + 1/4} = {2 \over 3}$.

A bit tedious, but it helps to go through the details. Also, intuitively it makes sense that the act of drawing a green marble increases the prior probability of the original marble being green.