$\Large \frac{\binom{6}{2} \times \binom{4}{2} }{\binom{10}{4}} = \frac{90}{210} = \frac{3}{7}$

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To select 2 red balls out of 6 red balls we can choose in ** $\binom{6}{2}$**. Similarly, for green balls we can select 2 red ball out of 4 red balls in **$\binom{4}{2}$.**

to get 4 balls out 10 balls, we can choose in $\binom{10}{4}$.

Therefore probability = **$\frac{\binom{6}{2} * \binom{4}{2}}{\binom{10}{4}}$= 3/7**

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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 $= (1/3)*(3/14)*6 = 3/7$

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 $= (1/3)*(3/14)*6 = 3/7$

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