A) \[\frac{15}{{{2}^{8}}}\]
B) \[\frac{2}{15}\]
C) \[\frac{15}{{{2}^{13}}}\]
D) None of these
Correct Answer: C
Solution :
[c] Let n be the number of tosses and X the number of times heads occurs. Then \[X\tilde{\ }B(n,p),\] with \[p=1/2\]. Therefore, since\[P(X=7)=P(X=9),\] we have \[^{n}{{C}_{7}}{{\left( \frac{1}{2} \right)}^{7}}{{\left( \frac{1}{2} \right)}^{n-1}}{{=}^{n}}{{C}_{9}}{{\left( \frac{1}{2} \right)}^{9}}{{\left( \frac{1}{2} \right)}^{n-9}}{{\Rightarrow }^{n}}{{C}_{7}}{{\left( \frac{1}{2} \right)}^{n}}{{=}^{n}}{{c}_{9}}{{\left( \frac{1}{2} \right)}^{n}}\]That is, \[^{n}{{C}_{7}}{{=}^{n}}{{C}_{9}}{{=}^{n}}{{C}_{n-9}},\] yielding \[7=n-9\]or \[n=16.\]Hence \[P(X=2){{=}^{16}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{16}}=\left( \frac{16\times 15}{2} \right){{\left( \frac{1}{2} \right)}^{16}}=\frac{15}{{{2}^{13}}}\]
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