Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$, then $\mathrm{k}$ is equal to :

Given that, <br/><br/>$\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times) <br/><br/>$$ \begin{aligned} & \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\\\ & \Rightarrow{ }^n \mathrm{C}_7={ }^n \mathrm{C}_9 \\\\ & \Rightarrow n=7+9=16 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Hence, } \mathrm{P}(\text { Even number twice })=\frac{k}{2^{15}} \\\\ & \Rightarrow{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2}=\frac{k}{2^{15}} \\\\ & \Rightarrow \frac{16 \times 15}{2} \times \frac{1}{2^{16}}=\frac{k}{2^{15}} \Rightarrow k=60 \end{aligned} $$