what is the probability that at least one of the marbles drawn is green? I thinks it's 1/11 but not sure.

Then what is the probability that only one of the marbles drawn are green?

You need to repeat the data on the post you are sending to get an answer.

To find the probability that at least one of the marbles drawn is green, you need to first determine the total number of outcomes and the number of favorable outcomes.

Let's assume you are selecting marbles from a bag that contains red, blue, and green marbles. To find the probability, you need to know the number of green marbles, the total number of marbles, and the number of marbles you're going to draw.

Let's say there are 5 green marbles, 3 red marbles, and 3 blue marbles in the bag, and you are drawing 2 marbles.

To find the probability that at least one of the marbles drawn is green, you can calculate the complement of the event that no green marbles are drawn. In other words, find the probability that all the marbles drawn are not green, and subtract it from 1.

The probability of not drawing a green marble in the first draw would be (8 non-green marbles) / (11 total marbles). After the first draw, since one marble has been drawn, there are now 10 marbles left in the bag, with 7 non-green marbles. So, the probability of not drawing a green marble in the second draw would be (7 non-green marbles) / (10 remaining marbles).

To find the probability that no green marbles are drawn, you multiply the probabilities of each draw:
(8/11) * (7/10) = 56/110 = 28/55

Therefore, the probability of drawing at least one green marble is:
1 - (28/55) = 27/55.

So, the correct probability that at least one of the marbles drawn is green is 27/55, not 1/11.

Now, the probability that only one of the marbles drawn is green can be calculated by considering two cases:
1) The first marble drawn is green, and the second marble drawn is not green.
2) The first marble drawn is not green, and the second marble drawn is green.

For the first case, the probability of drawing a green marble followed by a non-green marble would be:
(5/11) * (6/10) = 30/110 = 3/11

For the second case, the probability of drawing a non-green marble followed by a green marble would be:
(6/11) * (5/10) = 30/110 = 3/11

To get the total probability of only one marble being green, you add the probabilities of both cases:
(3/11) + (3/11) = 6/11.

So, the probability that only one of the marbles drawn is green is 6/11.