A fair coin is tossed a fixed number of times.
Question:

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. (1) $\frac{15}{2^{12}}$

  2. (2) $\frac{15}{2^{13}}$

  3. (3) $\frac{15}{2^{14}}$

  4. (4) $\frac{15}{2^{8}}$


Correct Option: , 2

Solution:

$\mathrm{p}(\mathrm{x}=9)=\mathrm{p}(\mathrm{x}=7)$

${ }^{\mathrm{n}} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{\mathrm{n}-9} \times\left(\frac{1}{2}\right)^{9}={ }^{\mathrm{n}} \mathrm{C}_{7}\left(\frac{1}{2}\right)^{\mathrm{n}-7} \times\left(\frac{1}{2}\right)^{7}$

${ }^{\mathrm{n}} \mathrm{C}_{9} \times\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2} \times{ }^{\mathrm{n}} \mathrm{C}_{7}$

$\mathrm{x}+\mathrm{y}=\mathrm{n} \quad \Rightarrow \mathrm{n}=16$

$\mathrm{p}(\mathrm{x}=2)={ }^{16} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{14} \times\left(\frac{1}{2}\right)^{2}$

$={ }^{16} \mathrm{C}_{2} \times\left(\frac{1}{2}\right)^{16}=\frac{15}{2^{13}}$

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