What is the probability of picking two green marbles without replacement?

The probability of picking two green marbles without replacement depends on the number of green marbles and total marbles.

To calculate this probability, you need to know the total number of marbles in the bag and how many of those marbles are green. Let's say there are \( G \) green marbles and \( T \) total marbles. When you pick the first marble, the probability that it is green is \( \frac{G}{T} \).

After removing one green marble, there are now \( G-1 \) green marbles left and \( T-1 \) total marbles left. The probability that the second marble you pick is also green is \( \frac{G-1}{T-1} \).

To find the overall probability of both events happening (picking two green marbles in a row), you multiply the probabilities of each individual event. So, the probability of picking two green marbles without replacement is:

\[ \frac{G}{T} \times \frac{G-1}{T-1} \]

For example, if you have a bag with 5 green marbles and 10 total marbles, the probability of picking the first green marble is \( \frac{5}{10} \) or \( \frac{1}{2} \). After removing one green marble, there are 4 green marbles left out of 9 total marbles. The probability of picking a second green marble is \( \frac{4}{9} \). Therefore, the combined probability is:

\[ \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9} \]

So, the probability of picking two green marbles without replacement in this example is \( \frac{2}{9} \).