A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be green?

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

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

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

$=\dfrac{90}{210}$

$=\dfrac{3}{7}$