A) \[\frac{1}{8}\]
B) \[\frac{3}{8}\]
C) \[\frac{1}{2}\]
D) \[\frac{2}{3}\]
Correct Answer: C
Solution :
Required probability \[={{\left( \frac{1}{2} \right)}^{3}}.{}^{3}{{C}_{2}}+{{\left( \frac{1}{2} \right)}^{3}}.{}^{3}{{C}_{3}}=\frac{4}{8}=\frac{1}{2}.\] {Here at least two heads means two heads or three heads}.
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