So here is the problem,
There are 3 boxes and each box contains 3 coins. The first one has three golden coins, the second a golden one and two silvers and the third one has three silver coins. If we select a box randomly and it has a golden coin, what is the possibility that we will drag a golden coin from the same box again?
I tried to solve the problem like this:
$B_1=\{\text{the event of chosing the first box}\}$
$B_2=\{\text{the event of choosing the second box}\}$
$B_3=\{\text{the event of choosing the third box}\}$
$G=\{\text{the event of choosing a golden coin}\}$
$GG=\{\text{the event of choosing two golden coins in a row}\}$
We are actually looking for $$P(GG|G)=\frac{P(GG \cap G)}{P(G)}$$
So,
$P(G) = P(B_1)\times P(G_1)+P(B_2)\times P(G_2)+P(B_3)\times P(G_3)= \frac{1}{3}\times 1 + \frac{1}{3}\times \frac{1}{3} + \frac{1}{3}\times 0 = \frac{4}{9}$ (There are basically three different ways this can happen and I multiply the events leading to a golden coin in each way)
$P(GG) = P(B_1)\times P(G_1) \times P(GG_1)+P(B_2)\times P(G_2) \times P(GG_2)+P(B_3)\times P(G_3) \times P(GG_3)= \frac{1}{3}\times 1 \times 1 + \frac{1}{3}\times \frac{1}{3} \times 0 + \frac{1}{3}\times 0 \times 0= \frac{1}{3}$
(Again I multiply the events leading to the second golden coin considering that when we drag a golden coin, we don't replace it)
My difficulty here is calculating $P(GG \cap G)$
I tried $P(GG \cap G) = P(G)+P(GG) = \frac{4}{9} + \frac{1}{3} =\frac{7}{9}$
but $$P(GG|G)=\frac{\frac{7}{9}}{\frac{4}{9}}=\frac{7}{4}$$
Another attemp was $P(GG \cap G) = P(G|GG)P(GG) = \frac{1}{3} \times\frac{1}{3} =\frac{1}{9}$
and thus $$P(GG|G)=\frac{\frac{1}{9}}{\frac{4}{9}}=\frac{1}{4}$$ which intuitively feels correct but I don't understand the logic behind these two different approaches. Can someone please tell me if the second one is correct and why the first one isn't.
The event of drawing two gold coins in a row includes the event of drawing the first cold coin. That is, you cannot draw two gold coins without drawing that first gold coin. Put differently, the $G$ event is not constraining the $GG$ in any way. Hence:
$$P(GG\cap G) = P(GG)$$