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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 are red and two are green ?

  1. $\large\frac{3}{7}$
  2. $\large\frac{4}{7}$
  3. $\large\frac{5}{7}$
  4. $\large\frac{6}{7}$


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1 Answer

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$4$  balls can be chosen from  $10$  balls in   $^{10}C_4$ ways. (Total number of ways)

The desirable(favourable) case is choosing $2$ red balls from $6$ red balls and choosing $2$ green balls from $4$ green balls.

So the required probability would be:  $\frac{^{6}C_2\times ^{4}C_2}{^{10}C_4}$  $=$  $\frac{3}{7}$

Option A is correct.
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