A fair coin is tossed k times. If the probability of getting exactly six tails is equal to the probability of getting exactly 8 tails, then the probability of getting exactly 3 tails is:

$\frac{91}{2^{12}}$

The correct answer is Option (2) → $\frac{37}{2^{12}}$

$P(T=6) = P(T=8)$

$\frac{k!}{6!(k-6)!} = \frac{k!}{8!(k-8)!}$

$\frac{1}{6!(k-6)!} = \frac{1}{8!(k-8)!}$

$8!(k-8)! = 6!(k-6)!$

$8 \cdot 7 \cdot 6!(k-8)! = 6!(k-6)(k-7)(k-8)!$

$56 = (k-6)(k-7)$

$k^2 -13k +42 = 56$

$k^2 -13k -14 = 0$

$k = 14$

$= \frac{14 \cdot 13 \cdot 12}{6 \cdot 2^{14}} = \frac{364}{16384}$

$= \frac{91}{4096}$

$\text{Required probability} = \frac{91}{4096}$