We have $6$ Red & $4$ Green balls.
Probability of $2$ Balls out of $4$ being Red $=(6/10)*(5/9) = 1/3$
Probability of other $2$ Balls being Green$=(4/8)*(3/7) = 3/14$
Now They can occur in any of$ (4!)/(2!*2!) = 6$ ways i.e RRGG, RGRG, RGGR, GGRR, GRGR, GRRG.
So Total probability $= (1/3)*(3/14)*6 = 3/7$