A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is
Correct Option: 1
Let the coin be tossed n-times
$\mathrm{P}(\mathrm{H})=\mathrm{P}(\mathrm{T})=\frac{1}{2}$
$P(7$ heads $)={ }^{n} C_{7}\left(\frac{1}{2}\right)^{n-7}\left(\frac{1}{2}\right)^{7}=\frac{{ }^{n} C_{7}}{2^{n}}$
$\mathrm{P}(9$ heads $)={ }^{n} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{\mathrm{n}-9}\left(\frac{1}{2}\right)^{9}=\frac{{ }^{n} \mathrm{C}_{9}}{2^{n}}$
$P(7$ heads $)=P(9$ heads $)$
${ }^{\mathrm{n}} \mathrm{C}_{7}={ }^{\mathrm{n}} \mathrm{C}_{9} \Rightarrow \mathrm{n}=16$
$\mathrm{P}(2$ heads $)={ }^{16} \mathrm{C}_{2}\left(\frac{1}{2}\right)^{14}\left(\frac{1}{2}\right)^{2}=\frac{15 \times 8}{2^{16}}$
$\mathrm{P}(2$ heads $)=\frac{15}{2^{13}}$