# 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 :   Option: 1 $\frac{15}{2^{8}}$ Option: 2 $\frac{15}{2^{13}}$ Option: 3 $\frac{15}{2^{14}}$   Option: 4 $\frac{15}{2^{12}}$

Let the coin tossed n times

$\\\mathrm{P}(\mathrm{H})=\mathrm{P}(\mathrm{T})=\frac{1}{2} \\ \mathrm{P}(7 \text { heads })={ }^{\mathrm{n}} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{\mathrm{n}-7}\left(\frac{1}{2}\right)^{7}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{7}}{2^{\mathrm{n}}} \\ \mathrm{P}(9 \text { heads })={ }^{\mathrm{n}} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{\mathrm{n}-9}\left(\frac{1}{2}\right)^{9}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{9}}{2^{\mathrm{n}}}$

Given that

$\\ { }^{n} C_{7}={ }^{n} C_{9} \Rightarrow n=16 \\ P(2 \text { heads })={ }^{16} C_{2}\left(\frac{1}{2}\right)^{14}\left(\frac{1}{2}\right)^{2}=\frac{15 \times 8}{2^{16}} \\ P(2 \text { heads })=\frac{15}{2^{13}}$

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