This solution explains how to find the probability of randomly selecting two green balls from a box containing balls of different colors.
We need to find the total number of ways to choose any 2 balls from the 12 available. This is a combination problem, denoted as $C(n, k)$ or $\binom{n}{k}$, where $n$ is the total number of items, and $k$ is the number of items to choose.
Total possible outcomes = $C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66$
Next, we find the number of ways to choose exactly 2 green balls from the 6 green balls available.
Favorable outcomes = $C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability (both balls are green) = $\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}} = \frac{15}{66}$
To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is 3.
Probability (both balls are green) = $\frac{15 \div 3}{66 \div 3} = \frac{5}{22}$
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