A) \[\frac{1}{3}\]
B) \[\frac{2}{7}\]
C) \[\frac{1}{21}\]
D) \[\frac{2}{23}\]
Correct Answer: B
Solution :
Total balls = 3 red balls + 4 blue balls + 2 green balls = 9 balls required probability\[=\frac{^{3}{{C}_{1}}{{\times }^{4}}{{C}_{1}}{{\times }^{2}}{{C}_{1}}}{^{9}{{C}_{3}}}=\frac{2}{7}\]
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