A bag contains 6 red balls, 4 green balls and 10 blue balls. Three balls are drawn with replacement. The probability of getting at least 1 green ball is:

The bag contains:

• Red balls = 6
• Green balls = 4
• Blue balls = 10
• Total balls = 20

Three balls are drawn with replacement, meaning after each draw the ball is returned to the bag.

For "at least 1" problems, use the complement rule:

$P(\text{at least 1 green}) = 1 - P(\text{no green balls})$

To find the probability of getting no green balls, the balls drawn must be either red or blue.

Non-green balls = 6 + 10 = 16 balls

For each draw:

$P(\text{not green}) = \frac{16}{20} = \frac{4}{5}$

Since the draws are independent (with replacement):

$P(\text{no green in all 3 draws}) = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}$

$P(\text{no green in all 3 draws}) = \left(\frac{4}{5}\right)^3$

$P(\text{no green in all 3 draws}) = \frac{64}{125}$

Using the complement rule:

$P(\text{at least 1 green}) = 1 - \frac{64}{125}$

$P(\text{at least 1 green}) = \frac{125}{125} - \frac{64}{125}$

$P(\text{at least 1 green}) = \frac{61}{125}$

Therefore, the probability of getting at least 1 green ball is $\frac{61}{125}$.