A) \[\frac{3}{8}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
Since both heads and tails appears, so \[n(S)=\{HHT,\,HTH,\,THH,\,HTT,\,THT,\,TTH\}\] \[n(E)=\{HTT,\,THT,\,TTH\}\] Hence required probability \[=\frac{3}{6}=\frac{1}{2}.\]
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