A coin is tossed K times. If the probability of getting 3 heads is equal to the probability of getting 7 heads, then the probability of getting 8 tails is:

$\frac{45}{1024}$

The correct answer is Option (3) → $\frac{45}{1024}$

$P(3\text{ heads})=P(7\text{ heads}).$

$\frac{K!}{3!(K-3)!}\left(\frac{1}{2}\right)^K=\frac{K!}{7!(K-7)!}\left(\frac{1}{2}\right)^K.$

$\frac{1}{3!(K-3)!}=\frac{1}{7!(K-7)!}.$

$7!(K-7)!=3!(K-3)!.$

$\frac{7!}{3!}=\frac{(K-3)!}{(K-7)!}.$

$\frac{7\cdot6\cdot5\cdot4}{1}=(K-3)(K-4)(K-5)(K-6).$

$840=(K-3)(K-4)(K-5)(K-6).$

$K=10.$

$\text{Probability of 8 tails }=P(2\text{ heads}).$

$=\frac{10!}{2!8!}\left(\frac{1}{2}\right)^{10}.$

$=\frac{45}{1024}.$

$\text{Required probability}=\frac{45}{1024}.$