The problem here involves calculating the probability of drawing two green marbles (denoted as $P(GG)$) from a bag that contains two green marbles and one yellow marble. The process involves two consecutive picks, with the first marble being replaced before the second pick.

Let's break this down step-by-step:

1. Initial Bag Setup:

   • The bag contains two green (G) marbles and one yellow (Y) marble.
   • So, the probability of picking a green marble on the first draw is $P(G_1) = \frac{2}{3}$ since there are 2 green marbles out of 3 total marbles.

2. After the First Pick:

   • Since the marble is replaced, the probabilities remain the same for the second pick.
   • The probability of picking a green marble on the second draw, given that the first was green, is still $P(G_2) = \frac{2}{3}$.

3. Calculation of $P(GG)$:

   • Since the events are independent (because the marble is replaced), we multiply the probabilities of drawing green on the first and second picks: $P(GG) = P(G_1) \times P(G_2) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}.$

Thus, the probability of drawing two green marbles is $P(GG) = \frac{4}{9}$.

Would you like more details on the calculation or have any other questions?

Related Questions:

1. What is the probability of picking a yellow marble on both picks $P(YY)$?
2. What is the probability of picking one green and one yellow marble $P(GY \text{ or } YG)$?
3. How does the scoring system affect the game’s strategy?
4. What would happen if there were more marbles of each color in the bag?
5. How would the probabilities change if the marbles were not replaced after being picked?

Tip: When events are independent, the probability of both happening is the product of their individual probabilities!