The question asks for the probability of selecting two green balls when drawing two balls randomly from a box containing balls of different colors. We need to calculate the number of ways to choose 2 green balls versus the total number of ways to choose any 2 balls.
First, find the total number of balls in the box:
Total balls = (Number of black balls) + (Number of green balls) + (Number of yellow balls)
Total balls = 2 + 6 + 4 = 12
Calculate the total number of ways to choose 2 balls from the 12 available balls. This uses combinations, denoted as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
Total combinations = $\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66$
Calculate the number of ways to choose 2 green balls from the 6 green balls available:
Favorable combinations = $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$
The probability of an event is the ratio of favorable outcomes to total possible outcomes:
Probability (both green) = $\frac{\text{Favorable combinations}}{\text{Total combinations}}$
Probability (both green) = $\frac{15}{66}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
Probability (both green) = $\frac{15 \div 3}{66 \div 3} = \frac{5}{22}$
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