Problem:
There are 4 red balls, 5 white balls, and 3 green balls in a basket. Three balls are chosen at random. What is the probability that there is at most 1 green ball?

Solution:
To find the probability of at most 1 green ball, we need to calculate the probability of selecting 0 green balls and the probability of selecting exactly 1 green ball.

Probability of selecting 0 green balls:
To calculate this probability, we need to consider that there are a total of 12 balls in the basket. So, the probability of selecting a non-green ball on the first draw is 9/12 (since there are 9 non-green balls out of the total 12 balls).

After the first draw, there are 11 balls left in the basket, out of which 8 are non-green. So, the probability of selecting a non-green ball on the second draw is 8/11.

Similarly, after the second draw, there are 10 balls left in the basket, out of which 7 are non-green. So, the probability of selecting a non-green ball on the third draw is 7/10.

To find the probability of selecting 0 green balls, we multiply the individual probabilities of each draw:
P(0 green balls) = (9/12) * (8/11) * (7/10)

Probability of selecting exactly 1 green ball:
To calculate this probability, we need to consider the different ways in which we can select exactly 1 green ball out of the 3 draws.

Case 1: Green ball on the first draw
The probability of selecting a green ball on the first draw is 3/12 (since there are 3 green balls out of the total 12 balls).

After the first draw, there are 11 balls left in the basket, out of which 4 are green. So, the probability of selecting a non-green ball on the second draw is 8/11.

Similarly, after the second draw, there are 10 balls left in the basket, out of which 7 are non-green. So, the probability of selecting a non-green ball on the third draw is 7/10.

To find the probability of selecting exactly 1 green ball in this case, we multiply the individual probabilities of each draw:
P(1 green ball, case 1) = (3/12) * (8/11) * (7/10)

Case 2: Green ball on the second draw
The probability of selecting a non-green ball on the first draw is 9/12.

After the first draw, there are 11 balls left in the basket, out of which 3 are green. So, the probability of selecting a green ball on the second draw is 3/11.

Similarly, after the second draw, there are 10 balls left in the basket, out of which 7 are non-green. So, the probability of selecting a non-green ball on the third draw is 7/10.

To find the probability of selecting exactly 1 green ball in this case, we multiply the individual probabilities of each draw:
P(1 green ball, case 2) = (9/12) * (3/11) * (7/10)

Case 3: Green ball on the third