A) \[\frac{1}{2}\]
B) \[\frac{1}{8}\]
C) \[\frac{3}{8}\]
D) None of these
Correct Answer: A
Solution :
The total number of cases are \[{{2}^{100}}\] The number of favourable ways \[{}^{100}{{C}_{1}}+{}^{100}{{C}_{3}}+\,.......\,+{}^{100}{{C}_{99}}={{2}^{100-1}}={{2}^{99}}\] Hence required probability \[=\frac{{{2}^{99}}}{{{2}^{100}}}=\frac{1}{2}.\]
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