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:

A fair coin is tossed $n$ times. The probability of getting 11 heads equals the probability of getting 13 heads.

For $n$ coin tosses, the probability of getting exactly $k$ heads is:

$P(k \text{ heads}) = C(n,k) \times \left(\frac{1}{2}\right)^n$

where $C(n,k) = \frac{n!}{k!(n-k)!}$

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Since $P(11 \text{ heads}) = P(13 \text{ heads})$:

$C(n,11) \times \left(\frac{1}{2}\right)^n = C(n,13) \times \left(\frac{1}{2}\right)^n$

$C(n,11) = C(n,13)$

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Using the property $C(n,r) = C(n, n-r)$:

$C(n,11) = C(n,13)$ implies $11 = n - 13$

$n = 24$

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The probability of getting 2 heads when $n = 24$:

$P(2 \text{ heads}) = C(24,2) \times \left(\frac{1}{2}\right)^{24}$

$C(24,2) = \frac{24!}{2! \times 22!}$

$C(24,2) = \frac{24 \times 23}{2 \times 1}$

$C(24,2) = \frac{552}{2}$

$C(24,2) = 276$

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$P(2 \text{ heads}) = \frac{276}{2^{24}}$

Simplifying where $276 = 2^2 \times 69$:

$P(2 \text{ heads}) = \frac{2^2 \times 69}{2^{24}}$

$P(2 \text{ heads}) = \frac{69}{2^{22}}$