Here total number of balls = 6 + 12 = 18
Two balls from 18 can be drawn in18C2 = \(\frac{18 \times 17}{1\times 2} = 153\)
One red ball out of 6 red can be drawn in 6C1 = 6 ways.
One green balls from 12 green may be done in 12C1 = 12 ways.
Therefore, number of favorable cases = 6 × 12 = 72
The probability that one is red and other is green = \(\frac{72}{153} = \frac{8}{17}\).