Question

an urn contains balls in three colors: blue, red, and green. there are 20 balls in the urn, 9 of which are blue, 7 of which are red, and 4 of which are green. if four balls are randomly chosen from the urn without replacement, what is the probability that at least one of them is green?

Answer 1

The probability of drawing at least one green ball is approximately 67.02%.

Step-by-step explanation:

To find the probability of drawing at least one green ball, we need to calculate the probability of not drawing any green balls and subtract it from 1.

There are a total of 20 balls in the urn. The probability of not drawing a green ball on the first draw is 16/20 (since there are 16 non-green balls out of 20 in total).

Since we draw the balls without replacement, the probability of not drawing a green ball on the second, third, and fourth draws are 15/19, 14/18, and 13/17 respectively. Therefore, the probability of not drawing any green balls is (16/20) * (15/19) * (14/18) * (13/17) ≈ 0.3298.

Finally, the probability of drawing at least one green ball is 1 - 0.3298 = 0.6702, or approximately 67.02%.