Solution

Total balls are

$6+4=10$

Number of ways to select $4$ balls from $10$ balls is

${}^{10}C_4$

We need exactly $2$ red balls and $2$ green balls.

Number of ways to select $2$ red balls from $6$ red balls is

${}^{6}C_2$

Number of ways to select $2$ green balls from $4$ green balls is

${}^{4}C_2$

So, required probability is

$\dfrac{{}^{6}C_2 \times {}^{4}C_2}{{}^{10}C_4}$

$=\dfrac{15 \times 6}{210}$

$=\dfrac{90}{210}$

$=\dfrac{3}{7}$