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: , 2
$\mathrm{p}(\mathrm{x}=9)=\mathrm{p}(\mathrm{x}=7)$
${ }^{\mathrm{n}} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{\mathrm{n}-9} \times\left(\frac{1}{2}\right)^{9}={ }^{\mathrm{n}} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{\mathrm{n}-7} \times\left(\frac{1}{2}\right)^{7}$
${ }^{\mathrm{n}} \mathrm{C}_{9} \times\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2} \times{ }^{\mathrm{n}} \mathrm{C}_{7}$
$\mathrm{x}+\mathrm{y}=\mathrm{n} \quad \Rightarrow \mathrm{n}=16$
$\mathrm{p}(\mathrm{x}=2)={ }^{16} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{14} \times\left(\frac{1}{2}\right)^{2}$
$={ }^{16} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{16}=\frac{15}{2^{13}}$
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