A) \[\frac{1}{2}\]
B) \[1-\frac{^{30}{{C}_{15}}}{{{2}^{30}}}\]
C) \[\frac{1}{2}+\frac{^{30}{{C}_{15}}}{{{2}^{31}}}\]
D) \[\frac{1}{2}+\frac{^{30}{{C}_{15}}}{{{2}^{30}}}\]
Correct Answer: C
Solution :
[c] According to the question, required probability \[P(E)=\frac{1}{{{2}^{30}}}{{[}^{30}}{{C}_{15}}{{+}^{30}}{{C}_{16}}{{+}^{30}}{{C}_{17}}+...{{+}^{30}}{{C}_{30}}]\] ? (1) Or \[P(E)=\frac{1}{{{2}^{30}}}{{[}^{30}}{{C}_{0}}{{+}^{30}}{{C}_{1}}{{+}^{30}}{{C}_{2}}+...{{+}^{30}}{{C}_{15}}]\] ? (2) Adding (1) and (2), we get \[2P(E)=\frac{1}{{{2}^{30}}}[{{(}^{30}}{{C}_{0}}{{+}^{30}}{{C}_{1}}+....{{+}^{30}}{{C}_{15}}{{+}^{30}}{{C}_{\grave{\ }16}}+...\] \[{{+}^{30}}{{C}_{30}}){{+}^{30}}{{C}_{15}}]\] \[=\frac{1}{{{2}^{30}}}[{{2}^{30}}{{+}^{30}}{{C}_{15}}]=1+\frac{^{30}{{C}_{15}}}{{{2}^{30}}}\] \[\therefore \,\,\,P(E)=\frac{1}{2}+\frac{^{30}{{C}_{15}}}{{{2}^{31}}}\]
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