A) \[\frac{5}{{{2}^{20}}}\]
B) \[\frac{7}{{{2}^{20}}}\]
C) \[\frac{4}{{{2}^{20}}}\]
D) \[\frac{6}{{{2}^{20}}}\]
Correct Answer: A
Solution :
\[S=\{1,2,...,20\}\] Total no. of subsets \[={{2}^{20}}\] We have,\[1+2+...+20=\frac{20(21)}{2}=210\] Sum of elements will be 203 if we would not choose 7 or elements which are giving sum 7. Elements giving sum 7 are (1, 6), (2, 5), (3, 4), (1, 2, 4) Thus, a subset will be 'nice' if we didn?t choose these 5 outcomes. \[\therefore \]Required probability (subset chosen is 'nice') \[=\frac{5}{{{2}^{20}}}\]
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