Conditional probability regarding drawing a marble This problem is Introduction to Probability (2019 2 edn), Chapter 2, Exercise 22, p 8:

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?

Here's how I tried to reason.
Define,
$A$: Event that the first marble drawn out is green.
$B$: Event that the second marble drawn out is green (computing the probability of this remaining marble is unaffected by the action of drawing it out).
Then, what we seek is $P(B|A)$, which by the definition of conditional probability, equals $\dfrac{P(A \cap B)}{P(B)}$. This is well-defined since we know that $B$ is not an impossible event.
Now, the sample space for the experiment of drawing out the two marbles is $\{GG,GB, BG\}$. Hence, $P(A \cap B) = \frac{1}{3}$. Also, $P(B) = \frac{2}{3}$. And so, $P(A|B) = \frac{1}{2}$.
I haven't read the solution to the problem, but a quick glimpse at the final answer tells me that my above answer is wrong. But I can't seem to find the flaw in the above reasoning. Could someone tell where I'm going wrong (hopefully, without revealing the solution to the problem)?
 A: The outcomes in the sample space $\{GG,GB, BG\}$ are not equally likely. The probability that we observe $GB$ and $BG$ are each $1/4$ and the probability that we observe $GG$ is $1/2$.
A: I feel that this problem can be solved intuitively, without calculating through Bayes' Rule.
Consider the following possible scenarios:


*

*The first marble taken out is green (given). This marble is the initial marble in the bag (assumed for the sake of this scenario). Then, the final draw will reveal a green marble.

*The first marble taken out it is green (given). This marble is the second marble put into the bag (assumed for the sake of this scenario). The initial marble is green (assumed for the sake of this scenario). Then, the final draw will reveal a green marble.

*The first marble taken out is green (given). This marble is the second marble put into the bag (assumed for the sake of this scenario). The initial marble is blue (assumed for the sake of this scenario). Then, the final draw will reveal a blue marble.


It seems that these are the only possible scenarios, and 2 out of 3 have the remaining marble as green. So P(remaining marble is green) = 2/3.
Verify with LOTP and Bayes'.
Let: A be the event that initial marble is green, B be the event that the first marble taken out is green, and C be the event that the remaining marble is green.
LOTP : $P(C|B) = P(C|B,A)P(A|B) +P(C|B,A^{c})P(A^{c}|B)$ = $P(A|B)$
Bayes': $$P(A|B) =\frac{P(B|A)P(A)}{P(B)}$$
$$P(A|B) =\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}$$
$$P(A|B) = \frac{1*\frac{1}{2}}{1*\frac{1}{2}+\frac{1}{2}*\frac{1}{2}}=\frac{2}{3}$$
