A fair coin is tossed for a fixed number of times. If probability of getting 7 heads is equal to probability of getting 9 heads, then probability of getting 2 heads is

\(\mathrm{p}=\) probability of getting head \(=\frac{1}{2}\) and
\(\mathrm{q}=\) probability of not getting head \(\frac{1}{2}\)
\(\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{n}-\mathrm{x}}\)
As per data given, we write
\(\begin{aligned}
& \mathrm{P}(\mathrm{x}=7)=\mathrm{P}(\mathrm{x}=9) \\
& \therefore{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{\mathrm{n}-7}={ }^{\mathrm{n}} \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{\mathrm{n}-9} \\
& \therefore \quad{ }^{\mathrm{n}} \mathrm{C}_7={ }^{\mathrm{n}} \mathrm{C}_9 \\
& \therefore \quad \mathrm{n}=16
\end{aligned}\)
when \(x=2\), we get
\(\mathrm{P}(\mathrm{x}=2)={ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{16 \times 15}{2}\left(\frac{1}{2}\right)^{16}=\frac{15}{(2)^{13}}\)