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A box contains 3 orange balls,3 green balls and blue balls.three balls are drawn at random from the box without replacement .The probability of drawing 2 green balls and one blue ball is

$\begin{array}{1 1}(A)\;\large\frac{3}{28}\\(B)\;\large\frac{2}{21}\\(C)\;\large\frac{1}{28}\\(D)\;\large\frac{167}{168}\end{array} $

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1 Answer

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  • three ball drawn from 3O,3G,2B balls without replacement
  • we can get 2 Green and 1Blue in the following ways
  • (G G B) (G B G) (B G G)
p(G G B) =\(\frac{3}{8}\)X\(\frac{2}{7}\)X\(\frac{2}{6}\)
p(G B G)=\(\frac{3}{8}\)X\(\frac{2}{7}\)X\(\frac{2}{6}\)
p(B G G )=\(\frac{2}{8}\)X\(\frac{3}{7}\)X\(\frac{2}{6}\)
p(getting 2Green and 1 Blue)
=\(\frac{9}{84}\)=\(\frac{3}{28}\)      ,A, option is correct
answered Feb 20, 2013 by poojasapani_1
edited Feb 20, 2013 by poojasapani_1

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