A fair coin is tossed a fixed number of times. If the probability of getting 11 heads is equal to the probability of getting 13 heads, then the probability of getting 2 heads is:

$\frac{69}{2^{22}}$

The correct answer is Option (1) → $\frac{69}{2^{22}}$

Let the coin be tossed $n$ times. Probability of $k$ heads: $P(k) = \frac{n!}{k!(n-k)!} \left(\frac{1}{2}\right)^n$

Given: $P(11) = P(13)$

$\frac{n!}{11!(n-11)!} = \frac{n!}{13!(n-13)!}$

$13!(n-13)! = 11!(n-11)!$

$12 \cdot 13 \cdot (n-13)! = (n-12)(n-11)(n-13)!$

$156 = (n-12)(n-11)$

$n^2 - 23n - 24 = 0$

$n = 24$ (positive solution)

Probability of 2 heads:

$P(2) = \frac{24!}{2!(24-2)!} \left(\frac{1}{2}\right)^{24} = \frac{24 \cdot 23}{2} \left(\frac{1}{2}\right)^{24} = 276 \cdot \frac{1}{2^{24}} = \frac{69}{2^{22}}$