A fair coin is tossed


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

  1. $\frac{15}{2^{13}}$

  2. $\frac{15}{2^{12}}$

  3. $\frac{15}{2^{8}}$

  4. $\frac{15}{2^{14}}$

Correct Option: 1


Let the coin be tossed n-times


$P(7$ 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 $)=\frac{15}{2^{13}}$

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