A) \[\frac{2}{9}\]
B) \[\frac{7}{27}\]
C) \[\frac{1}{27}\]
D) None of these
Correct Answer: B
Solution :
Required probability\[=P\](exactly two)\[+P\](exactly three) \[={}^{3}{{C}_{2}}.{{\left( \frac{2}{6} \right)}^{2}}\left( \frac{4}{6} \right)+{}^{3}{{C}_{3}}{{\left( \frac{2}{6} \right)}^{3}}=\frac{2}{9}+\frac{1}{27}=\frac{7}{27}\].
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