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Question

# Mark the correct alternative in the following question: A box contains 3 orange balls, 3 green balls and 2 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 $\left(\mathrm{a}\right)\frac{167}{168}\left(\mathrm{b}\right)\frac{1}{28}\left(\mathrm{c}\right)\frac{2}{21}\left(\mathrm{d}\right)\frac{3}{28}$

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Solution

## $\mathrm{Let}:\phantom{\rule{0ex}{0ex}}O\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{orange}\mathrm{ball},\phantom{\rule{0ex}{0ex}}G\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{green}\mathrm{ball}\mathrm{and}\phantom{\rule{0ex}{0ex}}B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{drawing}\mathrm{a}\mathrm{blue}\mathrm{ball}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\mathrm{n}\left(O\right)=3,\mathrm{n}\left(G\right)=3\mathrm{and}\mathrm{n}\left(B\right)=2\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{total}\mathrm{balls}=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(\mathrm{drawing}2\mathrm{green}\mathrm{balls}\mathrm{and}\mathrm{one}\mathrm{blue}\mathrm{ball}\right)=\mathrm{P}\left(GGB\right)+\mathrm{P}\left(GBG\right)+\mathrm{P}\left(BGG\right)\phantom{\rule{0ex}{0ex}}=\mathrm{P}\left(G\right)×\mathrm{P}\left(G|G\right)×\mathrm{P}\left(B|GG\right)+\mathrm{P}\left(G\right)×\mathrm{P}\left(B|G\right)×\mathrm{P}\left(G|GB\right)+\mathrm{P}\left(B\right)×\mathrm{P}\left(G|B\right)×\mathrm{P}\left(G|BG\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{8}×\frac{2}{7}×\frac{2}{6}+\frac{3}{8}×\frac{2}{7}×\frac{2}{6}+\frac{2}{8}×\frac{3}{7}×\frac{2}{6}\phantom{\rule{0ex}{0ex}}=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}\phantom{\rule{0ex}{0ex}}=\frac{3}{28}$ Hence, the correct alternative is option (d).

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