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?
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Correct answer: B. 3/7
Explanation
P = C(6,2)·C(4,2)/C(10,4) = 15 × 6 / 210 = 90/210 = 3/7.
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